Chapter 19, Problem 19.20QP

(a)

Interpretation Introduction

Interpretation:

From the given information the following has to be determined.

1. (a) The order of the reaction has to be calculated.

Concept introduction:

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

• Depends on order of the chemical reaction, the rate law or rate equation also varies.

Answer to Problem 19.20QP

The order of the reaction is third order

Explanation of Solution

To determine the order of the reaction

From the given information, experiments 2 and 5 denotes that when you double the concentration of $\text{X}$ at constant concentration of $\text{Y}$, the rate quadruples.

Let us take the ratio of these two rates,

$\begin{array}{l}\frac{rate5}{rate2}=\frac{0.509M/s}{0.127M/s}\approx 4=\frac{k{\left(0.40\right)}^x{\left(0.30\right)}^y}{k{\left(0.20\right)}^x{\left(0.30\right)}^y}\\ Therefore,\\ \frac{k{\left(0.40\right)}^x}{k{\left(0.20\right)}^x}=2^x=4\end{array}$

Or, $\text{x = 2}$.  Why because, the reaction is second order in $\text{X}$

From the given information, experiments 2 and 4 denotes that doubling $\text{[Y]}$ at constant $\text{[X]}$ doubles the rate.

Let us take the ratio of these two rates,

$\begin{array}{l}\frac{rate4}{rate2}=\frac{0.254M/s}{0.127M/s}\approx 2=\frac{k{\left(0.20\right)}^x{\left(0.60\right)}^y}{k{\left(0.20\right)}^x{\left(0.30\right)}^y}\\ Therefore,\\ \frac{k{\left(0.60\right)}^y}{k{\left(0.30\right)}^y}=2^y=2\end{array}$

Or, $y=1$. Why because the reaction is first order in $\text{Y}$

Hence, form the rate law

${\text{rate = k[X]}}^{\text{2}}\text{[Y]}$

$\text{The order of the reaction is}$ $\text{ (2 + 1) = 3}$

Therefore the order of the reaction is third order.

(b)

Interpretation Introduction

Interpretation:

From the given information the following has to be determined.

1. (b) The order of the reaction has to be calculated.

1. (c) The initial rate of disappearance has to be calculated.

Concept introduction:

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

• Depends on order of the chemical reaction, the rate law or rate equation also varies.

Answer to Problem 19.20QP

The initial rate of disappearance is $0.38M/s$

Explanation of Solution

To determine the initial rate of disappearance

From the given data, the rate constant k is calculated by rearranging the rate law.

$k=\frac{rate}{{\text{[X]}}^{\text{2}}\text{ [Y]}}=\frac{0.053}{(0.10M)2(0.50M)}=10.6M^{-2}{s}^{-1}$

Finally, by substituting the corresponding values in rate law and then by doing some mathematical calculation we get final answer

$\text{rate = (10}{\text{.6 M}}^{\text{-2}}.s^{\text{-1}}){(0.30M)}^2(0.40M)= 0.38M/s$