Question

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected: Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s) 1 0.50 0.50 0.50 1.5×10−4 2 0.50 0.50 1.50 4.5×10−4 3 1.00 0.50 0.50 6.0×10−4 4 1.00 1.00 0.50 6.0×10−4 Rate law equation The rate of a chemical reaction depends on the concentrations of the reactants. For the general reaction between Aand B, aA+bB⇌cC+dD The dependence of the reaction rate on the concentration of each reactant is given by the equation called the rate law: rate=k[A]m[B]n where k is a proportionality constant called the rate constant. The exponent m determines the reaction order with respect to A, and n determines the reaction order with respect to B. The overall reaction order equals the sum of the exponents (m+n).

Part D What is the value of the rate constant k for this reaction? When entering compound units, indicate multiplication of units explicitly using a multiplication dot (multiplication dot in the menu). For example, M−1⋅s−1. Express your answer to two significant figures and include the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot or a dash.

Answer 1

Solution:- Let's first write the given data in the correct format as..

Trial [A](M) [B](M) [C](M) Initial rate (M/s)

1. 0.50 0.50 0.50 1.5 x 10^-4

2. 0.50 0.50 1.50 4.5 x 10^-4

3. 1.00 0.50 0.50 6.0 x 10^-4

4. 1.00 1.00 0.50 6.0 x 10^-4

Let's say the order of reaction with respect to A is m, with respect to B is n and with respect to C is p. The rate law could be written as...

rate = k [A]^m [B]ⁿ [C]^p

we will make two equations and use those two trials where the concentrations for two of the reactants are constant and only one of them is changing to find out the order of reaction for each of the reactant.

In trial 1 and 2 only the concentration of C is changing while A and B are constant. we will use these two trials to find out the order of reaction with respect to C.

1.5 x 10^-4 = k[0.50)^m (0.50)ⁿ (0.50)^p   --------------(1)

4.5 x 10^-4 = k (0.50)^m (0.50)ⁿ (1.50)^p   --------------(2)

on dividing first equation by second equation...

1/3 = (1/3)^p

p = 1

using trial 1 and trial 3 data we could find out the value of m.

1.5 x 10^-4 = k[0.50)^m (0.50)ⁿ (0.50)^p   --------------(1)

6.0 x 10^-4 = k (1.0)^m (0.50)ⁿ (0.50)^p   ----------------(3)

dividing equation 1 by 3..

1/4 = (1/2)^m

(1/2)² = (1/2)^m

m = 2

now let's use trial 3 and 4 to find out the value of n.

6.0 x 10^-4 = k (1.0)^m (0.50)ⁿ (0.50)^p   ----------------(3)

6.0 x 10^-4 = k (1.0)^m (1.0)ⁿ (0.50)^p   ------------- (4)

dividing equation 3 by 4....

1 = (1/2)ⁿ

taking log to both sides..

log (1) = n log(1/2)

0 = n(-0.3010)

n = 0

from the above calculations, the order of reaction with respect to A is 2, with respect to B is 0 and with respect to C is 1 so the rate law would be written as..

rate = k [A]² [C]

over all order of reaction = 2+1 = 3

to find out the value of k we will use any of the given trial values and plug in it into the rate law.

1.5 x 10^-4 M/s = k[0.50 M)² (0.50 M)   

1.5 x 10^-4 M/s = k (0.50 M)³

k = (1.5 x 10^-4 M/s)/(0.50 M)³

k = 1.2 x 10^-3 M^-2.s^-1