Chapter 10, Problem 89CP

Interpretation Introduction

Interpretation:

The enthalpy change for the following reaction should be determined.

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{N}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Concept Introduction:

Enthalpy of a reaction is a state function thus; change in enthalpy of a reaction does not depend on the path of the reaction.

The change in enthalpy is same if the same reaction takes place in one step or series of steps. This is known as Hess’s law. There are following characteristics of change in enthalpy which are important to calculate the change in enthalpy of a reaction using Hess’s law.

1. For a reversed reaction, the sign of enthalpy change of the reaction also reversed.
2. If any reactant or product is multiplied by any integer in the chemical reaction, enthalpy change of that reactant or product is also multiplied by the same integer.

Answer to Problem 89CP

$-622.5\text{ kJ}$.

Explanation of Solution

Given Information:

The following reactions are given:

$\begin{array}{l}2{\text{NH}}_{\text{3}}\left(\text{g}\right)+3{\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)\to 4{\text{N}}_{\text{2}}\left(\text{g}\right){\text{+3H}}_2\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-1010\text{ kJ }\\ {\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-317\text{ kJ}\\ {\text{ 2NH}}_{\text{3}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to {\text{N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-143\text{ kJ}\\ {\text{H}}_{\text{2}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-\text{286 kJ}\end{array}$

Calculation:

The net reaction is as follows:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{N}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

To obtain the above reaction,

Reverse the 2^nd reaction as follows:

${\text{N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\to {\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\text{Δ}\text{H}=+317\text{ kJ}$

Multiply the above reaction by 2

${\text{2N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+2H}}_2\text{O}\left(\text{l}\right)\to 2{\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)+6{\text{H}}_{\text{2}}\left(\text{g}\right)\text{Δ}\text{H}=+634\text{ kJ}$

Adding the above reaction to 1^st reaction thus,

$\begin{array}{l}{\text{3N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+3H}}_2\text{O}\left(\text{l}\right)\to 3{\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)+9{\text{H}}_{\text{2}}\left(\text{g}\right)\text{Δ}\text{H}=+951\text{ kJ}\\ \underset{\_}{2{\text{NH}}_{\text{3}}\left(\text{g}\right)+3{\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)\to 4{\text{N}}_{\text{2}}\left(\text{g}\right){\text{+3H}}_2\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-1010\text{ kJ}}\\ {\text{3N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right)+2{\text{NH}}_{\text{3}}\left(\text{g}\right)\to 4{\text{N}}_{\text{2}}\left(\text{g}\right)+9{\text{H}}_{\text{2}}\left(\text{g}\right)\text{Δ}\text{H}=-59\text{ kJ}\end{array}$

Now, reverse the 3^rd reaction:

${\text{N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\to {\text{2NH}}_{\text{3}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\text{Δ}\text{H=+143 kJ}$

Adding this reverse reaction to above net reaction:

$\begin{array}{l}{\text{3N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right)+2{\text{NH}}_{\text{3}}\left(\text{g}\right)\to 4{\text{N}}_{\text{2}}\left(\text{g}\right)+9{\text{H}}_{\text{2}}\left(\text{g}\right)\text{Δ}\text{H}=-59\text{ kJ}\\ \underset{\_}{{\text{N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\to {\text{2NH}}_{\text{3}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\text{Δ}\text{H}=\text{+143 kJ}}\\ {\text{4N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\to 4{\text{N}}_{\text{2}}\left(\text{g}\right)+9{\text{H}}_{\text{2}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\text{Δ}\text{H}=\text{+84 kJ}\end{array}$

The 4^th reaction is multiplied by 9 thus,

${\text{9H}}_{\text{2}}\left(\text{g}\right)+\frac{9}{2}{\text{O}}_2\left(\text{g}\right)\to 9{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-\text{2574 kJ}$

Adding this to the above net reaction:

$\begin{array}{l}{\text{9H}}_{\text{2}}\left(\text{g}\right)+\frac{9}{2}{\text{O}}_2\left(\text{g}\right)\to 9{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-\text{2574 kJ}\\ \underset{\_}{{\text{4N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\to 4{\text{N}}_{\text{2}}\left(\text{g}\right)+9{\text{H}}_{\text{2}}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\text{Δ}\text{H}=\text{+84 kJ}}\\ {\text{4N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right)+4{\text{O}}_2\left(\text{g}\right)\to 4{\text{N}}_{\text{2}}\left(\text{g}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-\text{2490 kJ}\end{array}$

Divide the resultant reaction by 4 to get the desired reaction:

${\text{N}}_{\text{2}}{\text{H}}_4\left(\text{l}\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{Δ}\text{H}=-622.5\text{ kJ}$.