Chemical Reactions and Stoichiometry
Oxidation and Reduction in Chemical Reactions
Definition of Oxidation and Reduction (Redox Reactions):
Oxidation: The loss of electrons by an atom, ion, or molecule, resulting in an increase in its oxidation state.
Reduction: The gain of electrons by an atom, ion, or molecule, resulting in a decrease in its oxidation state.
Redox reactions always occur simultaneously; one species is oxidized while another is reduced.
Determine if the highlighted reactant has been oxidized or reduced in each of the following reactions:
2AgBr(s) \rightarrow 2Ag(s) + Br_{2}(g)
In AgBr, the silver ion is Ag^{+1} and the bromide ion is Br^{-1}.
Ag+ (oxidation state +1) in AgBr is reduced to Ag (oxidation state 0) by gaining one electron.
Br- (oxidation state -1) in AgBr is oxidized to Br2 (oxidation state 0) by losing electrons.
Conclusion: Within the reactant AgBr, the silver ion (Ag+) is reduced, while the bromide ion (Br-) is oxidized. This reaction is a decomposition redox reaction.
2S(s) + 3O{2}(g) \rightarrow 2SO{3}(g)
Sulfur (S) as an element has an oxidation state of 0.
In SO3, oxygen typically has an oxidation state of -2. For SO3 to be neutral, the sulfur must have an oxidation state of +6 (S + 3(-2) = 0 \Rightarrow S = +6).
S (oxidation state 0) is oxidized to S in SO3 (oxidation state +6) (losing electrons).
O2 (oxidation state 0) is reduced to O in SO3 (oxidation state -2) (gaining electrons).
Conclusion: Sulfur (S) \rightarrow Oxidized.
2C{2}H{2}(g) + 5O{2}(g) \rightarrow 4CO{2}(g) + 2H_{2}O(g)
This is a combustion reaction.
In acetylene (C2H2), hydrogen typically has an oxidation state of +1. For C2H2, 2C + 2(+1) = 0 \Rightarrow 2C = -2 \Rightarrow C = -1.
In CO2, oxygen has an oxidation state of -2. For CO2, C + 2(-2) = 0 \Rightarrow C = +4.
C in acetylene (oxidation state -1) is oxidized to C in CO2 (oxidation state +4) (increased oxidation state, losing electrons).
O2 (oxidation state 0) is reduced to O in CO2 and H2O (oxidation state -2) (gaining electrons).
Conclusion: Carbon (C) from C2H2 \rightarrow Oxidized.
Combustion Reaction
What is always the oxidizing agent in a combustion reaction?
A combustion reaction is a high-temperature exothermic redox chemical reaction between a fuel (reductant) and an oxidant, usually atmospheric oxygen, to produce oxidized, often gaseous products.
a. Oxygen (O2)
Explanation: In most combustion reactions, molecular oxygen (O2) acts as the primary oxidizing agent. An oxidizing agent (or oxidant) is a substance that causes another substance to lose electrons (get oxidized) and is itself reduced in the process. Oxygen accepts electrons from the fuel, leading to the oxidation of the fuel while oxygen atoms are reduced from an oxidation state of 0 (in O2) to -2 (in combustion products like CO2 and H2O).
b. Water
c. Heat
d. Alkane
Conclusion: The correct answer is a. Oxygen. As the primary electron acceptor, oxygen is the oxidizing agent in typical combustion reactions.
Catalytic Hydrogenation
Catalytic Hydrogenation Process:
A chemical reaction between molecular hydrogen (H_2) and another compound, typically an unsaturated organic compound containing multiple bonds (double or triple carbon-carbon bonds, or carbon-heteroatom double bonds like C=O or C=N).
Involves hydrogen gas (H2) acting as a reducing agent; it donates electrons and is itself oxidized (from 0 to +1).
This process is greatly facilitated by the presence of a catalyst, typically a transition metal like platinum (Pt), palladium (Pd), or nickel (Ni).
Mechanism: The catalyst provides a surface where both hydrogen and the unsaturated compound can adsorb. This weakens the H-H bond and the multiple bond, allowing hydrogen atoms to add across the multiple bond, saturating it. This typically occurs in a syn-addition manner.
Applications: Important in various industrial processes, such as the production of saturated fats from unsaturated oils (e.g., margarine), the synthesis of pharmaceuticals, and the petrochemical industry.
Biological Reactions
Similar reduction-oxidation reactions take place extensively in biological systems, often catalyzed by specific enzymes. They are fundamental to many metabolic processes, including energy production (cellular respiration) and biosynthesis.
Example Reaction: C=C-C-S^{-} + NADPH + H^{+} \rightarrow C-C-C-S^{-} + NADP^{+}
This reaction exemplifies the reduction of a carbon-carbon double bond within a biological molecule, often seen in fatty acid synthesis.
NADPH (Nicotinamide Adenine Dinucleotide Phosphate in its reduced form) acts as a crucial biological reducing agent. It carries electrons (and a proton) and donates them to the substrate, becoming oxidized to NADP+.
The double bond (C=C) is reduced to a single bond (C-C) by the addition of hydrogen atoms.
This highlights the critical biological importance of reduction-oxidation processes in cellular metabolism, where electron transfer facilitates energy conversion and the synthesis of complex molecules.
Mole and Mass Relationships in Reactions
A balanced chemical equation provides perpetual, quantitative insights into chemical reactions, serving as a powerful predictive tool.
Example Reaction: 2H{2}(g) + O{2}(g) \rightarrow 2H_{2}O(g)
Information provided by a balanced equation:
Stoichiometric Ratio of Reactants to Products: The coefficients in the balanced equation represent the mole ratios in which reactants combine and products are formed. For this example, 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H_2O.
Conservation of Mass and Atoms: It demonstrates that matter is conserved; the number of atoms of each element remains constant before and after the reaction. Two H2 molecules and one O2 molecule transform into two H2O molecules, ensuring all hydrogen and oxygen atoms are accounted for.
Changes that occur during the reaction: It shows the chemical transformation from reactants to products.
Utility: Balancing chemical equations is essential as it allows for:
Prediction of Limiting Reactant: Identifying which reactant will be completely consumed first, thereby stopping the reaction.
Calculation of Theoretical Yields: Determining the maximum amount of product that can possibly be formed from a given set of reactants.
Calculation of Reactant Amounts: Determining how much of one reactant is needed to react completely with a given amount of another.
Stoichiometry in Reactions
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. These relationships are based on the law of conservation of mass and the concept of balanced chemical equations.
Practical application involving calculation: The principles of stoichiometry can be illustrated with everyday analogies.
Example: Bread and cheese slices calculation for making sandwiches.
If a sandwich requires 2 slices of bread and 1 slice of cheese, this represents a fixed "stoichiometric" ratio.
Given 20 slices of bread, calculate needed cheese slices.
Formula: 20 \text{ bread slices} \times \frac{1 \text{ cheese slice}}{2 \text{ bread slices}} = 10 \text{ cheese slices}.
This shows how to use a ratio (derived from the "recipe" or balanced equation) to determine the quantity of one component based on the quantity of another.
Molar Relationships in Reaction Calculations
Example Calculation: This demonstrates how to use molar mass and mole ratios from a balanced equation to convert between moles and grams.
Given reaction: 4KO{2}(s) + 2CO{2}(g) \rightarrow 2K{2}CO{3}(s) + 3O_{2}(g)
Problem: How many grams of KO2 are needed for 0.400 mol CO2?
Step 1: Determine moles of KO2 required using the stoichiometric mole ratio from the balanced equation.
From the equation, 4 \text{ mol KO}2 react with 2 \text{ mol CO}2.
0.400 \text{ mol CO}2 \times \frac{4 \text{ mol KO}2}{2 \text{ mol CO}2} = 0.800 \text{ mol KO}2
Step 2: Calculate grams of KO2 needed by multiplying the moles of KO2 by its molar mass.
The molar mass of KO2 (Potassium Superoxide) is calculated from the atomic masses: K (\approx 39.10 \text{ g/mol}) + 2 \times O (\approx 16.00 \text{ g/mol}) = 39.10 + 32.00 = 71.10 \text{ g/mol}.
Formula: 0.800 \text{ mol KO}2 \times \frac{71.10 \text{ g KO}2}{1 \text{ mol KO}2} = 56.88 \text{ g KO}2 (Rounded to three significant figures, this is 56.9 \text{ g KO}_2).
Moles and Ratios in Predictive Calculations
Example Calculation: Applying stoichiometric ratios to determine the amount of a reactant needed.
For reaction: 2 Al + 3 Cl{2} \rightarrow 2 AlCl{3}
Determination needed: How many moles of chlorine (Cl_2) are required if 0.250 moles of Aluminum (Al) are used?
To solve this, use the mole ratio from the balanced chemical equation. From the equation, 2 moles of Al react with 3 moles of Cl2.
Calculation: 0.250 \text{ mol Al} \times \frac{3 \text{ mol Cl}2}{2 \text{ mol Al}} = 0.375 \text{ mol Cl}2
Options:
a. 0.009 moles
b. 0.167 moles
c. 0.250 moles
d. 0.375 moles
Conclusion: Based on the stoichiometric calculation, the correct answer is d. 0.375 moles. This demonstrates the direct application of mole ratios in predicting reagent consumption.
Yield Calculations in Reactions
Definition of Limiting Reactant: The reactant that is completely consumed during a chemical reaction, thereby limiting the amount of product that can be formed. Once the limiting reactant is used up, the reaction stops, regardless of how much of the other reactants (excess reactants) are present.
Example Reaction: N{2}(g) + 3H{2}(g) \rightarrow 2NH_{3}(g) (Haber-Bosch process)
Given: 2.10 mol N2 and 5.70 mol H2.
To find the limiting reactant:
Method 1: Calculate how much of one reactant is needed to react with the given amount of the other.
Calculate total H2 needed to react completely with 2.10 mol N2:
2.10 \text{ mol N}2 \times \frac{3 \text{ mol H}2}{1 \text{ mol N}2} = 6.30 \text{ mol H}2Compare with available H2 (5.70 mol): Since 6.30 \text{ mol H}2 is needed but only 5.70 \text{ mol H}2 is available, H2 is the limiting reactant because there isn't enough of it to react with all the N2.
Method 2: Calculate how much product each reactant could make. The reactant that yields the least product is the limiting reactant.
From N2: 2.10 \text{ mol N}2 \times \frac{2 \text{ mol NH}3}{1 \text{ mol N}2} = 4.20 \text{ mol NH}3
From H2: 5.70 \text{ mol H}2 \times \frac{2 \text{ mol NH}3}{3 \text{ mol H}2} = 3.80 \text{ mol NH}3
Since H2 produces less NH3 (3.80 mol vs 4.20 mol), H2 is the limiting reactant.
The other reactant, N2, is the excess reactant.
General Reaction Equivalence Analysis
Statement analysis on a neutralization reaction involving Hydrochloric Acid (HCl) and Potassium Hydroxide (KOH):
Reaction: HCl + KOH \rightarrow KCl + H_{2}O
This is an acid-base reaction, not a redox reaction, as the oxidation states of H (+1), Cl (-1), K (+1), and O (-2) remain unchanged throughout the reaction. Therefore, there are no oxidizing or reducing agents in the traditional redox sense.
Given: 0.125 moles HCl and 0.300 moles KOH.
The stoichiometric ratio between HCl and KOH is 1:1. This means 1 mole of HCl reacts completely with 1 mole of KOH.
True statement options:
a. HCl is the reducing agent: False. Not a redox reaction.
b. KOH is the reducing agent: False. Not a redox reaction.
c. 0.150 moles HCl are needed: False. This amount is not directly supported by the stoichiometry with the given KOH.
d. 0.300 moles HCl are needed: This statement would be True if it refers to the amount of HCl required to completely react with the given 0.300 moles of KOH. Since the mole ratio is 1:1, 0.300 moles of KOH would require 0.300 moles of HCl for complete neutralization.
Analysis of Limiting/Excess Reactant for the Given Amounts:
Given 0.125 moles HCl and 0.300 moles KOH.
Since 1 mole HCl reacts with 1 mole KOH:
0.125 moles HCl will react with 0.125 moles KOH.
HCl is the limiting reactant.
KOH is the excess reactant (0.300 - 0.125 = 0.175 \text{ mol KOH} in excess).
Conclusion: Given the options, and assuming the question implies a stoichiometric need for complete reaction with the larger amount of one reactant: If we have 0.300 moles of KOH, then 0.300 moles of HCl are needed to react completely with it. Thus, d. 0.300 moles HCl are needed (to react completely with the available KOH) is the most plausible true statement focusing on a complete reaction with the stated amount of KOH.