Lecture 5
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ANY GENERAL QUESTIONS? Feel free to see me after class!
Practice Problems
- What mass of CaCl2 should be dissolved in 25.0 kg water so that the solution freezes at –1.50 °C? Assume 100% dissociation. For water, Kf = 1.86 °C/m. Answer: 746 g
- What amount of benzene (C6H6) is needed to dissolve 2.10 mol I2(s) to form a solution that boils at 83.7 °C? For benzene, the normal boiling point is 80.1 °C and Kb = 2.53 °C/m. Answer: 19 mol
- What is the osmotic pressure at 25 °C of an aqueous methanol (CH3OH) solution that boils at 102.52 °C? The density of the solution is 0.963 g/mL. For water, Kb = 0.512 °C/m. Answer: 100. atm
- Challenging: At 80 °C, 2.00 mol HF was added to 20.0 mol water, resulting in a solution with a vapor pressure of 314.3 mmHg. The vapor pressure of pure water at 80 °C is 355.1 mmHg. What percentage of HF dissociated? Answer: 30 %
Reaction Rate Problem
- Consider the reaction: N2(g) + 3 H2(g) \rightarrow 2 NH_3(g)
- A sealed 1.00-L container initially contains 5.00 moles N2 and 5.00 moles H2. After 25.0 s, 1.50 moles NH_3(g) is formed.
- Calculations:
- [NH3] = \frac{1.50 \, mol \, NH3}{1.00 \, L} = 1.50 \, M \, NH_3
- Moles of H2 consumed: 1.50 \, M \, NH3 \times \frac{3 \, M \, H2}{2 \, M \, NH3} = 2.25 \, M \, H_2
- Initial [H2] = 5.00 \, M, Final [H2] = 5.00 - 2.25 = 2.75 \, M
- a) Rate of change of [H_2]:
- \frac{\Delta [H_2]}{\Delta t} = \frac{2.75 - 5.00 \, M}{25.0 - 0 \, s} = \frac{-2.25 \, M}{25.0 \, s} = -0.0900 \, M/s
- b) Generalized rate of the reaction:
- rate = -\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = -\frac{1}{3} \frac{-0.0900 \, M}{s} = +0.0300 \, M/s
- (Alternatively, using NH3: \frac{1}{2} \frac{\Delta [NH3]}{\Delta t})
The Effect of Concentration on Reaction Rate
- The effect of concentration on reaction rate is given by the reaction’s rate law
- Rate Law (AKA the differential rate law)
- A mathematical equation that expresses the relationship between the rate of a reaction and the concentration of reactant (and, in rare cases, the concentration of product).
- MUST be experimentally determined (!)
- The order(?!?) of the reactants CANNOT be established from the balanced coefficients.
- Only exception is when dealing with elementary reactions of a reaction mechanism (more about this later).
The (Differential) Rate Law
- A \rightarrow products
- General form of the rate law: rate = k [A]^n
- k = rate constant for the reaction
- [A]^n = the molar concentration of A raised to some power, n
- n is called the order of the reactant
- n commonly is 0, 1, or 2 though it could be a fraction and/or negative in rare cases
- General form of the rate law: rate = k [A]^n
- A + B \rightarrow products
- General form of the rate law: rate = k [A]^m[B]^n
- k = rate constant for the reaction
- [A]^m = the molar concentration of A raised to some power, m
- [B]^n = the molar concentration of B raised to some power, n
- General form of the rate law: rate = k [A]^m[B]^n
Reaction Order (n, m…)
- The order of a reactant reflects the rate’s sensitivity to changing that reactant’s concentration.
- Rates of reactions with high orders are much more sensitive to changes in concentration than rates of reactions with low orders.
- Example: O2(g) + 2 NO(g) \rightarrow 2 NO2(g)
- The rate law is experimentally determined to be rate = k [O_2][NO]^2
- The reaction is first-order with respect to O_2 (m = 1)
- The reaction is second-order with respect to NO (n = 2)
- The reaction is third-order overall (m + n = 1 + 2 = 3)
The Rate Constant k
- Depends on the specific reaction and the temperature.
- reaction \, rate \propto k
- Doubling k will double the rate.
- Units of k depend on the overall order of the reaction.
- Hypothetically, what are the units for a fourth-order rate constant? M^{-3} \cdot s^{-1}
Zero Order Reaction (n = 0)
- rate = k [A]^0
- rate = k
- Zero-order k usually has units of M/s (or M \cdot s^{-1})
- The reaction rate depends only on the value of k.
- The rate of a zero-order reaction is constant as the reaction proceeds.
- Many reactions occurring on surfaces are zero-order reactions.
First Order Reaction (n = 1)
- rate = k [A]^1
- First-order k usually has units of 1/s (or s^{-1})
- The rate is directly proportional to the reactant concentration.
- Doubling the concentration of reactant will double the rate.
- Decreasing the concentration by a factor of 3 will decrease the rate by a factor of 3.
- The rate of a first-order reaction decreases as the reaction proceeds since the concentration of reactant decreases.
Second Order Reaction (n = 2)
- rate = k [A]^2
- Second-order k usually has units of 1/M⋅s (or M^{-1} \cdot s^{-1})
- The rate is directly proportional to the square of the reactant concentration.
- Doubling the concentration of reactant will quadruple (i.e., 2^2) the rate.
- Tripling the concentration increases the rate by a factor of 9 (i.e., 3^2).
- The rate of a second-order reaction is more sensitive to changes in concentration than the rate of a first-order reaction.
Reaction Plots – Rate vs. Reactant Concentration (A → products)
- zero-order reaction: the rate of the reaction is constant as reactant is consumed
- first-order reaction: the rate of the reaction slows down linearly (i.e., at a constant rate)
- second-order reaction: the rate of the reaction slows down at a faster pace as reactant is consumed
Experimentally Determining Rate Laws and Reaction Orders
- Use “the method of initial rates”:
- Measure the initial rate of a reaction under a set of initial concentrations for the reactant(s).
- Change the initial concentration of only one reactant (keeping the concentrations of the other reactants constant) and measure the initial rate under the new conditions.
- Establish the order with respect to only that one reactant.
- Go back and repeat the process for all the other reactants until the order of each reactant is established.
- Once all of the orders are established, calculate the value of the rate constant using the data from any one set of experiments.
Method of Initial Rates – Example 1
- Consider the following data for the reaction: A \rightarrow products
- Establish the full rate law (including the value of k) for this reaction
- Experiment #:
- [A]o (M) - Initial Rate (M/s)
- 1 - 0.125 - 0.00347
- 2 - 0.325 - 0.0234
- Calculations:
- Experiment 1: rate_1 = k [A]^n
- Experiment 2: rate_2 = k [A]^n
- \frac{0.00347}{0.0234} = \frac{k [0.125]^n}{k [0.325]^n}
- 0.148 = [\frac{0.125}{0.325}]^n
- 0.148 = [0.385]^n
- log(0.148) = log(0.385^n)
- log(0.148) = n \cdot log(0.385)
- n = \frac{log(0.148)}{log(0.385)} \approx 2
- rate = k [A]^2
- 0.00347 = k (0.125)^2
- k = 0.222 \, M^{-1} \cdot s^{-1}
- Final rate law: rate = (0.222 \, M^{-1} \cdot s^{-1})[A]^2
Method of Initial Rates – Example 2
- Consider the following data for the reaction: A + B + C → products
- Establish the full rate law (including the value of k) for this reaction and determine the initial rate of this reaction when the initial concentration of each reactant is 0.950 M.
- Experiment #:
- [A]o (M) - [B]o (M) - [C]o (M) - Initial Rate(M/s)
- 1 - 0.500 - 0.100 - 2.10 - 0.600
- 2 - 0.500 - 0.100 - 4.20 - 2.40
- 3 - 0.500 - 0.300 - 2.10 - 1.80
- 4 - 1.25 - 0.100 - 4.20 - 2.40
- Observations:
- ↑ × 2.5 no change increasing [A] by a factor of 2.5 leads to no change in reaction rate; the reaction is zero order with respect to A