Question

The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (M) Initial Rate (10–3 M/s) 0.4 0.4 0.2 160 0.2 0.4 0.4 80 0.6 0.1 0.2 15 0.2 0.1 0.2 5 0.2 0.2 0.4 20 Using the initial-rate method, what is the order of the reaction with respect to C? a. zero-order b. first-order c. third-order d. second-order e. impossible to tell from the data given

Answer 1

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

$Rate = k[A]^{x}[B]^{y}[C]^{z}$

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

$\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}$

$\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}$

$3 = 3^{x} \\\\x =1$

Order w.r.t B : Use trials 2 and 5

$\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}$

$\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}$

$4 = 2^{y} \\\\y =2$

Order w.r.t C : Use trials 1 and 2

$\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}$

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

$\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}$

$1 = (0.5)^{z}$

z = 1

Therefore, order w.r.t C = 1