Energy Changes, Reaction Profiles, and Reaction Rates

Energy Changes in Chemical Reactions

Chemical Bonds and Energy Changes:
- All chemical reactions involve changes in energy as chemical bonds are broken in the reactants and new chemical bonds are formed in the products.
- Bond breaking is an endothermic process that always absorbs energy from the surroundings.
- Bond formation is an exothermic process that always releases energy into the surroundings.
- The strength of chemical bonds dictates the magnitude of energy transfer: stronger bonds require more energy to break and release more energy when formed.
- The overall energy change of a chemical reaction is determined by the net balance between the energy absorbed during bond breaking and the energy released during bond formation: $\Delta H = \text{energy absorbed (breaking bonds)} - \text{energy released (forming bonds)}$
Heat of Reaction ( $\Delta H$ ):
- Definition: Heat of reaction ( $\Delta H$ ) is the net change of chemical potential energy of the system.
- The value of $\Delta H$ indicates whether a reaction absorbs or releases energy overall, classifying the reaction as either exothermic or endothermic.
Exothermic Reactions:
- Definition: Exothermic reactions are reactions which transform chemical potential energy into thermal energy.
- In an exothermic reaction, chemical potential energy is converted into thermal energy, which is released to the surroundings, resulting in a net release of energy ( $\Delta H < 0$ ).
- There is a net release of thermal energy to the surroundings, causing the temperature of the reaction vessel and surroundings to increase.
- Because potential energy is released, the products have lower chemical potential energy than the reactants.

Exothermic reaction vessel releasing thermal energy to the surroundings

Endothermic Reactions:
- Definition: Endothermic reactions are reactions which transform thermal energy into chemical potential energy.
- In an endothermic reaction, thermal energy is absorbed from the surroundings and converted into chemical potential energy, resulting in a net absorption of energy ( $\Delta H > 0$ ).
- There is a net absorption of thermal energy from the surroundings, causing the temperature of the reaction vessel and surroundings to decrease.
- Because energy is absorbed and stored within the bonds of the products, the products have higher chemical potential energy than the reactants.

Endothermic reaction vessel absorbing thermal energy from the surroundings

Activation Energy and Reaction Energy Profiles

Key Concepts of Energy Profiles:
- Activation Energy ( $E_a$ ): The minimum energy required to start a chemical reaction, or the energy required to form the activated complex.
- Activated Complex: A high energy, unstable, temporary transition state between the reactants and the products located at the peak of the potential energy curve.
- Reaction Coordinate: The horizontal axis representing the progress of the reaction from reactants to products.
- Potential Energy Axis: The vertical axis representing the chemical potential energy of the chemical species in joules ( $\text{J}$ ) or kilojoules ( $\text{kJ}$ ).
Exothermic Energy Profile:
- The reactants start at a higher potential energy level than the products.
- The activation energy ( $E_a$ ) is the energy difference between the reactants and the activated complex at the peak of the curve.
- The energy released represents the downward drop in energy from the activated complex to the final products.
- More energy is released during product formation than is absorbed during reactant bond breaking, resulting in $\Delta H < 0$ .
- The reaction vessel temperature increases as net chemical potential energy transforms into thermal energy.

Endothermic Energy Profile:
- The reactants start at a lower potential energy level than the products.
- The activation energy ( $E_a$ ) is the large energy difference from the reactants up to the activated complex.
- The energy released from the activated complex down to the products is less than the initial energy absorbed to reach the peak.
- More energy is absorbed than released, resulting in $\Delta H > 0$ .
- The reaction vessel temperature decreases as thermal energy is converted into chemical potential energy.

Worked Example 1: Analysis of an Energy Profile Diagram:

1. Missing Axis Labels:
- Vertical axis: Potential energy ( $\text{J}$ or $\text{kJ}$
- Horizontal axis: Reaction coordinate (or Course of reaction)
2. Identification of Components:
- (a): Activation energy ( $E_a$ )
- (b): Energy released (from activated complex to products)
- (c): Heat of reaction ( $\Delta H$ )
- (d): Reactants
- (e): Activated complex
- (f): Products
- (i): Potential energy of reactants
- (ii): Potential energy of activated complex
- (iii): Potential energy of products
3. Reaction Classification:
- This curve represents an endothermic reaction.
- Explanation: The energy level of the products is higher than the energy level of the reactants ( $\Delta H > 0$ ), showing that more energy was absorbed to break bonds than was released when new bonds formed.
4. Calculation of Activation Energy:
- Problem Statement: The reaction has a net energy absorption of $75\,kJ$ . Calculate the activation energy if the bonds release $250\,kJ$ of energy when they form.
- Calculation: $\Delta H = E_a - \text{energy released}$ $75\,kJ = E_a - 250\,kJ$ $E_a = 75\,kJ + 250\,kJ = 325\,kJ$
- Worked Example 2: Catalysed vs. Uncatalysed Reaction Energy Profiles:

1. Potential Energy of Reactants (uncatalysed): $20\,kJ$
2. Potential Energy of Products (uncatalysed): $50\,kJ$
3. Potential Energy of Reactants (catalysed): $20\,kJ$
4. Potential Energy of Products (catalysed): $50\,kJ$
Conclusion from Questions 1 to 4: A catalyst has no effect on the initial chemical potential energy of the reactants or the final chemical potential energy of the products.
6. Potential Energy of Activated Complex (uncatalysed): $165\,kJ$
- Activation energy (uncatalysed): $E_a = 165\,kJ - 20\,kJ = 145\,kJ$
7. Potential Energy of Activated Complex (catalysed): $90\,kJ$
- Activation energy (catalysed): $E_a = 90\,kJ - 20\,kJ = 70\,kJ$
Conclusion from Questions 6 and 7: A catalyst lowers the potential energy of the activated complex, thereby reducing the activation energy required for the reaction to take place.
9. Heat of Reaction ( $\Delta H$ ) for Uncatalysed Reaction: $\Delta H = 50\,kJ - 20\,kJ = +30\,kJ$
10. Heat of Reaction ( $\Delta H$ ) for Catalysed Reaction: $\Delta H = 50\,kJ - 20\,kJ = +30\,kJ$
Conclusion from Questions 9 and 10: A catalyst does not change the heat of reaction ( $\Delta H$ ); the net energy absorbed or released remains identical.

Reaction Rates and Experimental Measurement Techniques

Defining Reaction Rate:
- Definition: Reaction rate is the change in concentration per unit time of either a reactant or product.
- Reaction rate indicates the speed at which reactants are converted into products:
- Fast reactions: Precipitation reactions, neutralization of strong acids with strong bases.
- Moderate reactions: Dissolution of zinc or calcium carbonate in hydrochloric acid.
- Slow reactions: Rusting of iron over weeks or months.
- Mathematical representation: $\text{Rate of reaction} = \frac{\Delta[\text{concentration}]}{\Delta\text{time}} = \frac{[\text{final concentration}] - [\text{initial concentration}]}{\Delta\text{time}}$
Qualitative vs. Quantitative Observations:
- Qualitative Change: Visible, non-numerical observations used to infer reaction speed (e.g., observing rapid bubbling, sudden color appearance, or cloudiness).
- Quantitative Change: Measurable physical quantities tracked over time using instruments:
- Measuring volume of gas produced over time using a gas syringe or inverted measuring cylinder.
- Measuring the decrease in mass of a system over time using a top-pan mass balance.
- Measuring color changes using a light meter or colorimeter.
- Measuring precipitate formation (turbidity) using a light meter.
- Measuring changes in acidity or basicity over time using a pH meter.
- All reaction rate determinations require measuring these changes against elapsed time.
Experimental Methods for Rate Measurement:
- Model Reaction: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$
- Method 1: Measuring Mass Decrease over Time:
- The reaction mixture is placed in a conical flask sitting on an electronic top-pan balance.
- A loose cotton wool plug is placed in the neck of the flask to prevent liquid droplets from splashing out while freely allowing gaseous $CO_2$ to escape into the atmosphere.
- As carbon dioxide escapes, the total mass of the system decreases.
- The mass of the system is recorded at regular time intervals, yielding a downward-sloping curve whose gradient represents the reaction rate.

Top-pan balance apparatus for measuring mass decrease during gas evolution

Method 2: Measuring Gas Volume over Time:
- The conical flask containing calcium carbonate and acid is sealed with a rubber stopper fitted with a delivery tube.
- The delivery tube leads into an inverted measuring cylinder filled with water and submerged in a water trough.
- As $CO_2$ gas is generated, it displaces the water downward (downward displacement of water).
- The volume of gas collected in the cylinder is recorded at regular intervals, generating an upward-sloping curve that plateaus once the limiting reagent is fully consumed.

Inverted measuring cylinder apparatus for measuring gas volume by water displacement

Graphical Analysis of Reaction Rates

Interpreting Rate Graphs:
- The gradient (slope) of an amount versus time graph directly reflects the reaction rate:
- Steep gradient $\rightarrow$ high reaction rate.
- Gentle gradient $\rightarrow$ low reaction rate.
- Horizontal line (gradient = 0) $\rightarrow$ reaction has stopped (rate is zero).
Average Rate vs. Instantaneous Rate:
- Average Rate: The overall rate of reaction calculated over an extended time interval between two distinct points on a curve: $\text{Average rate} = \frac{\Delta\text{measured quantity}}{\Delta\text{time}}$
- Instantaneous Rate: The precise rate of reaction at a single specific moment in time.
- Determined by drawing a tangent line to the curve at the chosen time point and calculating the gradient ( $\frac{\Delta y}{\Delta x}$ ) of the tangent line.

Instantaneous rate shown as a tangent line to the reaction progress curve

Worked Example: Average Rate Calculation:

Graph showing volume of hydrogen gas produced versus time

From the graph, the volume of $H_2$ gas increases from $0\,dm^3$ to $3.0\,dm^3$ over an elapsed time of $40\,s$ , where the curve reaches its maximum plateau.
Calculation: $\text{Average rate} = \frac{\Delta V}{\Delta t} = \frac{3.0\,dm^3 - 0\,dm^3}{40\,s - 0\,s} = 0.075\,dm^3\,s^{-1}$
Alternatively, expressed in cubic centimeters: $\text{Average rate} = \frac{3000\,cm^3}{40\,s} = 75\,cm^3\,s^{-1}$
- Rate Progression over Time:
At the start ( $t = 0$ ), the gradient is at its steepest because the concentration of reactant particles is at its maximum, leading to the highest frequency of collisions.
As the reaction proceeds, reactants are converted into products, causing the concentration of reactant particles to decrease.
With fewer reactant particles present, fewer collisions occur per unit time, resulting in a continuously decreasing reaction rate (decreasing gradient).
When one or more of the reactants is completely consumed, collisions between reactant particles cease, the gradient becomes zero, and the graph forms a horizontal plateau.
- Analysis of Stages on an S-Shaped Curve:
Stage A - B (Initial Stage): Often shows an initial slow rate or induction period where reagents mix, or an accelerating rate if the reaction is exothermic and generates heat that increases kinetic energy.
Stage B - C (Middle Stage): Rapid rate of reaction, which gradually tapers off as reactant concentrations decline.
Stage C - D (Final Stage): Horizontal plateau indicating that the limiting reagent has been completely used up and the reaction has stopped.
- Temperature Comparison on Gas Volume Curves:
When comparing a reaction run at high temperature versus low temperature with equal quantities of limiting reagent:
- High temperature: Curve has a significantly steeper initial gradient (faster rate) and reaches the completion plateau in a shorter time period.
- Low temperature: Curve has a gentler initial slope (slower rate) and takes longer to reach completion.
- Both curves plateau at the exact same final volume of gas if the moles of the limiting reactant are identical.

Gas production comparison over time between high and low temperatures

Collision Theory and the Maxwell–Boltzmann Distribution

Core Principles of Collision Theory:
- A chemical reaction can only occur when reacting particles collide with one another.
- Not all collisions result in a chemical reaction; only effective collisions (successful collisions) lead to the breaking of existing bonds and the formation of new bonds.
- For a collision to be effective, two mandatory conditions must be simultaneously satisfied:
1. Correct Orientation: Particles must collide in the correct spatial alignment so that the specific reactive atoms contact each other and allow bond rearrangement.
2. Sufficient Kinetic Energy: Particles must possess kinetic energy equal to or greater than the activation energy ( $E_k \ge E_a$ ) to overcome the repulsive electron cloud forces and break bonds.
- If colliding particles have insufficient kinetic energy, or if they collide with incorrect orientation, they simply bounce apart unchanged with no reaction.

Particle collisions illustrating correct and incorrect orientation and energy

The Maxwell–Boltzmann Energy Distribution:
- In any sample of matter, particles exhibit a wide distribution of kinetic energies due to random collisions.
- Temperature is defined as a measure of the average kinetic energy of the particles ( $E_k \propto T$ ).
- The Maxwell–Boltzmann distribution curve plots the number of molecules (vertical axis) against kinetic energy in joules (horizontal axis):
- The curve starts at the origin $(0,0)$ because no molecules have zero kinetic energy.
- The peak of the curve represents the most probable kinetic energy possessed by the greatest number of molecules.
- The curve is asymmetrical, trailing off to the right at high energies but never touching the horizontal axis (as there is no theoretical upper limit to particle energy).
- The total area under the curve represents the total number of particles in the system.

Maxwell-Boltzmann distribution showing activation energy and shaded reacting area

Activation Energy Threshold on the Distribution:
- The activation energy ( $E_a$ ) is marked as a vertical line on the horizontal kinetic energy axis.
- Particles to the left of $E_a$ possess kinetic energy less than the activation energy ( $E_k < E_a$ ) and cannot react when they collide.
- Particles to the right of $E_a$ possess kinetic energy equal to or greater than the activation energy ( $E_k \ge E_a$ ).
- The shaded area under the curve to the right of $E_a$ represents the fraction of particles capable of undergoing an effective collision.
- Reaction rate is directly proportional to the number of successful collisions per unit time.
- Any factor that increases the total collision frequency or increases the shaded area ( $E_k \ge E_a$ ) increases the rate of reaction.

Factors Affecting Reaction Rates

The Five Factors Affecting Reaction Rates:
1. Temperature of the reaction mixture.
2. Concentration of reactants (or pressure of gaseous reactants).
3. Surface area (state of division) of solid reactants.
4. Presence of a catalyst (or light in photochemical reactions).
5. Nature of the reacting substances.
1. Effect of Temperature:
- An increase in temperature increases the average kinetic energy of the particles ( $E_k \propto T$ ), causing particles to move faster.
- This exerts a two-fold effect on the reaction rate:
- A significantly greater fraction of particles have sufficient kinetic energy to overcome the activation energy barrier ( $E_k \ge E_a$ ).
- The faster moving particles collide more frequently, resulting in more collisions per unit time.
- As a consequence, there are significantly more effective collisions per unit time, leading to a higher reaction rate.
- Rule of Thumb: A $10\,^\circ\text{C}$ (or $10\,\text{K}$ ) increase in temperature approximately doubles the rate of a chemical reaction.
- A temperature increase of $30\,^\circ\text{C}$ increases the rate by approximately: $2 \times 2 \times 2 = 8\text{ times}$
- Impact on the Maxwell–Boltzmann Distribution:
- As temperature increases from $T_1$ to a higher temperature $T_2$ , the curve flattens and shifts to the right.
- The peak becomes lower and shifts toward higher kinetic energy values.
- The total area under the curve remains constant (same total number of particles).
- The shaded area to the right of the unchanged $E_a$ line increases substantially, demonstrating that far more molecules possess $E_k \ge E_a$ .
2. Effect of Concentration:
- Concentration refers to the number of solute particles per unit volume of solution.
- An increase in the concentration of reactants means that there are more particles per unit volume.
- Particles are situated closer together, resulting in more collisions per unit time (increased collision frequency).
- This results in more effective collisions per unit time, leading to a higher reaction rate.
- Proportionality:
- For a generic reaction $A + B \rightarrow AB$ , the reaction rate is proportional to the product of reactant concentrations.
- Doubling the concentration of reactant $A$ doubles the reaction rate.
- Doubling the concentrations of both reactants $A$ and $B$ increases the reaction rate fourfold ( $2 \times 2 = 4$ ).
- On a Maxwell–Boltzmann distribution, increasing concentration increases the total number of particles, which scales the entire curve upward while keeping the shape, peak position, and $E_a$ line unchanged.
3. Effect of Pressure in Gaseous Reactions:
- The pressure of a gas is inversely proportional to the volume of its container ( $P \propto \frac{1}{V}$ ).
- An increase in pressure (caused by a decrease in the volume of the reaction container) forces gaseous particles closer together.
- This results in more gas particles per unit volume.
- There are more collisions per unit time between reacting particles.
- This results in more effective collisions per unit time, leading to a higher reaction rate.
4. Effect of Surface Area (State of Division):
- Surface area applies to heterogeneous reactions involving at least one solid reactant.
- In a large solid lump, only the atoms or molecules on the outer surface can collide with other reactant particles; particles in the interior are shielded from contact.
- Breaking a solid into smaller pieces or grinding it into a fine powder dramatically increases the total exposed surface area without changing the overall mass.
- An increase in surface area means that there is a greater number of solid surface particles exposed to collisions.
- There are more collisions per unit time between reactants.
- This results in more effective collisions per unit time, leading to a higher reaction rate.

Magnesium solid lump versus divided particles showing exposed surface area

5. Effect of a Catalyst:
- Definition: A catalyst is a substance that increases the rate of a chemical reaction without undergoing any permanent chemical change itself and remains unchanged at the end of the reaction.
- A catalyst provides an alternative reaction pathway with a lower activation energy ( $E_a$ ).
- Mechanism via Collision Theory:
- The catalyst offers a reaction pathway of lower activation energy.
- Consequently, a larger fraction of particles possess sufficient kinetic energy ( $E_k \ge E_a$ ) to overcome the reduced energy barrier.
- This results in more effective collisions per unit time.
- This leads to a higher reaction rate.
- A catalyst does not alter the energy of reactants, the energy of products, or the enthalpy change ( $\Delta H$ ).
- Impact on the Maxwell–Boltzmann Distribution: The kinetic energy curve of the particles remains completely unchanged; instead, the vertical threshold line for $E_a$ shifts to the left, which dramatically increases the shaded area representing particles capable of reacting.
6. Effect of the Nature of Reacting Substances:
- Because chemical reactions require breaking existing bonds and making new bonds, the rate depends heavily on bond types and chemical reactivity (such as oxidising and reducing strengths).
- Reactions Between Ions in Aqueous Solution:
- Occur almost instantaneously.
- In aqueous solution, ionic compounds are already dissociated into free, mobile ions.
- No strong covalent bonds need to be broken before new bonds form.
- Oppositely charged ions attract each other through strong electrostatic forces, leading to high collision success rates.
- Reactions Involving Covalent Compounds:
- Tend to be substantially slower at room temperature.
- Existing strong covalent bonds within molecules must be broken before atom rearrangement can occur, requiring higher activation energies.

Quantitative Applications and Case Studies

Case Study 1: Copper and Dilute Nitric Acid:
- Balanced Chemical Equation: $3Cu(s) + 8HNO_3(aq) \rightarrow 3Cu(NO_3)_2(aq) + 2NO(g) + 4H_2O(l)$
- The resulting solution is blue due to the formation of hydrated copper(II) ions ( $Cu^{2+}$ ).
- Experimental Data Table:

Experiment	Mass of $Cu\,(g)$	Volume of $HNO_3\,(dm^3)$	Concentration of $HNO_3\,(mol\cdot dm^{-3})$	State of $Cu(s)$
A	$2.54$	$0.10$	$0.8$	Small pellets
B	$2.54$	$0.05$	$0.8$	Large chunks
C	$5.08$	$0.05$	$1.6$	Small pellets

1. Moles of Nitric Acid in Experiment A: $n(HNO_3) = c \times V = 0.8\,mol\cdot dm^{-3} \times 0.10\,dm^3 = 0.08\,mol$
2. Proof that Copper is in Excess in Experiment A:
- Calculate the initial moles of copper available (molar mass of $Cu = 63.5\,g\cdot mol^{-1}$ ): $n(Cu)_{\text{initial}} = \frac{m}{M} = \frac{2.54\,g}{63.5\,g\cdot mol^{-1}} = 0.04\,mol$
- According to the balanced equation, $3\,mol$ of $Cu$ reacts with $8\,mol$ of $HNO_3$ (mole ratio $3 : 8$ ).
- Determine the moles of $Cu$ required to react completely with all $0.08\,mol$ of $HNO_3$ : $n(Cu)_{\text{reacted}} = 0.08\,mol \times \frac{3}{8} = 0.03\,mol$
- Since $0.04\,mol$ of $Cu$ is provided and only $0.03\,mol$ is consumed, copper is present in excess by $0.01\,mol$ , confirming that $HNO_3$ is the limiting reagent.
3. Volume of $NO(g)$ Produced at STP in Experiment A:
- From the balanced equation, $8\,mol$ of $HNO_3$ produces $2\,mol$ of $NO(g)$ .
- Moles of $NO$ produced: $n(NO) = n(HNO_3) \times \frac{2}{8} = 0.08\,mol \times \frac{1}{4} = 0.02\,mol$
- Using standard molar gas volume at STP ( $V_m = 22.4\,dm^3\cdot mol^{-1}$ ): $V(NO) = n \times V_m = 0.02\,mol \times 22.4\,dm^3\cdot mol^{-1} = 0.448\,dm^3 = 448\,cm^3$
4. Comparison and Graphical Behavior of Experiments B and C:
- Experiment B Comparison:
  - Moles of $HNO_3$ : $n = c \times V = 0.8\,mol\cdot dm^{-3} \times 0.05\,dm^3 = 0.04\,mol$ .
  - Because it has half the moles of limiting reagent compared to Experiment A ( $0.04\,mol$ vs. $0.08\,mol$ ), it produces exactly half the final volume of $NO$ gas ( $0.01\,mol\,NO$ ).
  - The copper is in "large chunks" rather than "small pellets," meaning a smaller surface area.
  - Curve B has a much gentler initial gradient (slower rate) and plateaus at half the height of Curve A.
- Experiment C Comparison:
  - Moles of $HNO_3$ : $n = c \times V = 1.6\,mol\cdot dm^{-3} \times 0.05\,dm^3 = 0.08\,mol$ .
  - The total moles of limiting reagent are identical to Experiment A ( $0.08\,mol$ ), so Experiment C produces the exact same total volume of $NO$ gas ( $0.02\,mol\,NO$ ).
  - The acid concentration is twice as high ( $1.6\,mol\cdot dm^{-3}$ vs. $0.8\,mol\cdot dm^{-3}$ ), and the copper is in small pellets.
  - Curve C has a significantly steeper initial gradient (faster rate) and plateaus at the same height as Curve A, but reaches completion in less time.
- Case Study 2: Nickel and Hydrochloric Acid Reaction:
Balanced Chemical Equation: $Ni(s) + 2HCl(aq) \rightarrow NiCl_2(aq) + H_2(g)$
Hydrogen gas is collected via downward displacement of water over a beehive shelf.

Experimental setup for collecting hydrogen gas over water using a beehive shelf

Acid is the limiting reagent in both experiments, and the nickel is completely covered.
Experimental Parameters:

Experiment	Volume of $HCl\,(dm^3)$	Concentration of $HCl\,(mol\cdot dm^{-3})$	Temperature ( $^\circ\text{C}$ )
1	$0.4$	$0.04$	$50$
2	$0.6$	$0.04$	$40$

1. Two Conditions for an Effective Collision:
- Reactant particles must collide with correct orientation.
- Reactant particles must collide with sufficient kinetic energy ( $E_k \ge E_a$ ).
2. Explaining the Effect of Increased $[HCl]$ via Collision Theory:
- An increase in the concentration of $HCl$ results in more $H^+$ and $Cl^-$ particles per unit volume of solution.
- This leads to an increased frequency of collisions between acid particles and the nickel surface (more collisions per unit time).
- Consequently, there are more effective collisions per unit time, resulting in an increased rate of reaction.
3. Determination of Total Volume of $H_2$ Gas Collected in Experiment 1:
- The graph of Volume versus Time for Experiment 1 shows that the reaction reaches its plateau at $t = 16\,\text{min}$ .
- Given that the average rate of formation of $H_2$ is $12\,cm^3\cdot min^{-1}$ : $\text{Total volume} = \text{average rate} \times \Delta t = 12\,cm^3\cdot min^{-1} \times 16\,min = 192\,cm^3$
4. Calculation of Percentage Yield for Experiment 1:
- Molar volume of gas at $50\,^\circ\text{C}$ : $V_m = 26\,490\,cm^3\cdot mol^{-1}$ .
- Calculate the initial moles of limiting reagent $HCl$ : $n(HCl) = c \times V = 0.04\,mol\cdot dm^{-3} \times 0.4\,dm^3 = 0.016\,mol$
- From the stoichiometry ( $2\,mol\,HCl : 1\,mol\,H_2$ ), calculate theoretical moles of $H_2$ : $n(H_2)_{\text{theoretical}} = \frac{0.016\,mol}{2} = 0.008\,mol$
- Calculate theoretical volume of $H_2$ : $V_{\text{theoretical}} = n \times V_m = 0.008\,mol \times 26\,490\,cm^3\cdot mol^{-1} = 211.92\,cm^3$
- Calculate percentage yield: $\text{Percentage yield} = \frac{\text{actual volume}}{\text{theoretical volume}} \times 100 = \frac{192\,cm^3}{211.92\,cm^3} \times 100 = 90.60\%$
5. Comparative Curve Analysis for Experiment 2:
- Moles of limiting reagent $HCl$ in Experiment 2: $n(HCl) = c \times V = 0.04\,mol\cdot dm^{-3} \times 0.6\,dm^3 = 0.024\,mol$
- Theoretical moles of $H_2$ in Experiment 2: $n(H_2) = \frac{0.024\,mol}{2} = 0.012\,mol$
- Because $0.012\,mol$ is $1.5$ times the amount produced in Experiment 1 ( $0.008\,mol$ ), Experiment 2 produces $1.5$ times more hydrogen gas ( $192\,cm^3 \times 1.5 = 288\,cm^3$ at the same yield), plateauing at a substantially higher volume.
- The temperature in Experiment 2 is lower ( $40\,^\circ\text{C}$ compared to $50\,^\circ\text{C}$ ), meaning particles have lower average kinetic energy and move slower.
- Therefore, the initial reaction rate in Experiment 2 is slower; its graph starts with a gentler gradient, takes longer to level off, and eventually reaches a plateau $1.5$ times higher than Experiment 1 within the 64-minute period.