Question

Consider the following hypothetical reaction and experimental data:
$A+B\to C+D$
$T=273K$
(A) What is the order with respect to $A$?
(B) What is the order with respect to $B$?
(C) What is the rate equation?
(D) What is the overall order of the reaction?
(E) Calculate the rate constant.

• A. A. $x=2$; B. $y=0.5$; C. rate $=k{\left[A\right]}^2{\left[B\right]}^{0.5}$; D. $2.5$; E. $3.5$
• B. A. $x=1$; B. $y=0.5$; C. rate $=k{\left[A\right]}^1{\left[B\right]}^{0.5}$; D. $5.5$; E. $3.5$
• C. A. $x=0$; B. $y=1.5$; C. rate $=k{\left[A\right]}^2{\left[B\right]}^{0.5}$; D. $1.5$; E. $3.5$
• D. A. $x=2$; B. $y=0$; C. rate $=k{\left[A\right]}^0{\left[B\right]}^{0.5}$; D. $2.5$; E. $3.5$

Answer 1

The correct option is A A. $x=2$; B. $y=0.5$; C. rate $=k{\left[A\right]}^2{\left[B\right]}^{0.5}$; D. $2.5$; E. $3.5$

(A) From experiments 1 and 3, the initial concentration of B is kept constant whereas the initial concentration of A is doubled from 0.10 mol/L to 0.20 mol/L. The rate of the reaction is is increased four times 0.35 M/s to 0.140 M/s. Hence, the order of the reaction with respect to A is 2.
(B) From experiments 1 and 2, the initial concentration of A is kept constant whereas the initial concentration of B is increased four times from 1 mol/L to 4 mol/L. The rate of the reaction is doubled from 0.35 M/s to 0.70 M/s. Hence, the order of the reaction with respect to B is 0.5.
(C) The rate equation is rate $=k\left[A{]}^2\right[B{]}^{0.5}$
(D) The overall order of the reaction is the sum of the orders with respect to A and B. It is $2+0.5=2.5$
(E) Substituting data for first experiment in rate expression, we get
$0.35=k\times (0.10{)}^2\times 1^{0.5}$
$k=35M^{-1.5}{s}^{-1}$