Chapter 15.4, Problem 34PP

(a)

Interpretation Introduction

Interpretation:

The given reaction is to be divided into steps.

Concept introduction:

Enthalpy of formation is the energy required to form one mole of that substance.

Answer to Problem 34PP

$\Delta {{\rm H}}_{rxn}=\Delta {{\rm H}}_f(N{O}_2)-\Delta {{\rm H}}_f(NO)$

Explanation of Solution

The given reaction is:

$2NO+O_2\to 2N{O}_2$

Formation of $N{O}_2$ can be done as:

Formation of $NO$ : $N_2+O_2\to 2NO$

Formation of $N{O}_2$ : $N_2+2O_2\to 2N{O}_2$

Thus adding the equation second to reverse of first equation will give resultant equation.

(b)

Interpretation Introduction

Interpretation:

The given reaction is to be divided into steps.

Concept introduction:

Enthalpy of formation is the energy required to form one mole of that substance.

Answer to Problem 34PP

$\Delta {{\rm H}}_{rxn}=\Delta {{\rm H}}_f(H_{2}S{O}_4)-\Delta {{\rm H}}_f(S{O}_3)-\Delta {{\rm H}}_f(H_{2}O)$

Explanation of Solution

The given reaction is:

$S{O}_3+H_{2}O\to H_{2}S{O}_4$

Formation of $H_{2}S{O}_4$ : $H_2+S+2O_2\to H_{2}S{O}_4$

Formation of $S{O}_3$ : $S+\frac{3}{2}{O}_2\to S{O}_3$

Formation of $H_{2}O$ : $H_2+\frac{1}{2}{O}_2\to H_{2}O$

Thus adding the equation first to reverse of second and third equation will give resultant equation.