Chapter 10, Problem 10.109QA

Interpretation Introduction

To find:

The percent decrease in the total pressure of a sealed reaction vessel during the reaction between $H_2$ and $N_2$.

Answer to Problem 10.109QA

Solution:

There will be $13\%$ decrease in total pressure of a sealed reaction vessel during the reaction between $H_2$ and $N_2$.

Explanation of Solution

1) Concept:

According to ideal gas law, pressure is directly proportional to number of moles of gas.

$P=nRT$

i.e. $P\propto n$

Decrease in percent can be calculated by the following formula:

$Percent decrease=\frac{Initial-Final}{Initial}\times 100$

2) Formula:

i) $P=nRT$

ii) $Percent decrease=\frac{Initial-Final}{Initial}\times 100$

3) Given:

i) Initial Partial pressure of $H_2\left(g\right), P_{H_2\left(g\right)}=2.4 atm$

ii) Initial Partial pressure of $N_2\left(g\right)P_{N_2\left(g\right)}=3.6 atm$

4) Calculation:

Total pressure before reaction is

$P_{T_{initial}}=P_{H_2\left(g\right)}+P_{N_2\left(g\right)}=3.6 atm+2.4 atm=6 atm$

The balanced equation is

$3H_2\left(g\right)+N_2\left(g\right)\to 2{NH}_3\left(g\right)$

Half of the hydrogen gas reacted, therefore partial pressure of $H_2\left(g\right)$ after reaction is

$\frac{2.4 atm{ H}_2\left(g\right)}{2}=1.2 atm H_2\left(g\right)$

Since the mole ratio of $H_2:N_2$ is 3:1, the initial partial pressure of $N_2$ gas is calculated as

$1.2 atm H_2\left(g\right)\times \frac{N_2\left(g\right)}{3H_2\left(g\right)}=0.40 atm$

Partial pressure of $N_2\left(g\right)$ after reaction is

$3.6 atm-0.4 atm=3.2 atm N_2\left(g\right)$

The mole ratio of $H_2:{NH}_3$ is 3:2. Therefore, the partial pressure of ${NH}_3\left(g\right)$ after the reaction is

$1.2 atm H_2\left(g\right)\times \frac{{2NH}_3\left(g\right)}{3H_2\left(g\right)}=0.80 atm{ NH}_3\left(g\right)$

Total pressure after the reaction is

$P_{T_{final}}=P_{H_2\left(g\right)}+P_{N_2\left(g\right)}+P_{{NH}_3\left(g\right)}=3.2 atm+1.2 atm+0.8 atm=5.2 atm$

Percent decrease in pressure after the reaction is

$Percent decrease=\frac{Initial-Final}{Initial}\times 100$

$Percent decrease in total pressure=\frac{6 atm-5.2 atm}{6 atm }\times 100=13\%$

Therefore, there will be $13\%$ decrease in the total pressure of a sealed reaction vessel during the reaction between $H_2$ and $N_2$.

Conclusion:

The partial pressure of each gas after the reaction is calculated using the stoichiometry of the reaction at the constant volume.