Chapter 14, Problem 14.1P

(a)

Interpretation Introduction

Interpretation:

To develop expressions for mole fractions of reacting species as function of reaction coordinate for

  $4N{H}_3(g)+5O_2(g)\to 4NO(g)+6H_{2}O(g)$

Concept Introduction :

The mole fractions y_i of the species present are related to $\epsilon$ by

  $y_i=\frac{n_i}{n}=\frac{n_{i_0}+v_i\epsilon }{n_0+v\epsilon }$

Answer to Problem 14.1P

  $yN{H}_3=\frac{2-4\epsilon }{7+\epsilon }$

  $y{O}_2=\frac{5-5\epsilon }{7+\epsilon }$

  $yNO=\frac{4\epsilon }{7+\epsilon }$

  $y{H}_{2}O=\frac{6\epsilon }{7+\epsilon }$

Explanation of Solution

  $4N{H}_3(g)+5O_2(g)\to 4NO(g)+6H_{2}O(g)$

  $v_i$ = Moles of products − Moles of reactants

  $v=v_i=4+6-4-5=1$

  $n_0=2mol(N{H}_3)+5mol(O_2)=7$

The mole fractions y_i of the species present are related to $\epsilon$ by

  $y_i=\frac{n_i}{n}=\frac{n_{i_0}+v_i\epsilon }{n_0+v\epsilon }$

  $yN{H}_3=\frac{2-4\epsilon }{7+\epsilon }$

  $y{O}_2=\frac{5-5\epsilon }{7+\epsilon }$

  $yNO=\frac{4\epsilon }{7+\epsilon }$

  $y{H}_{2}O=\frac{6\epsilon }{7+\epsilon }$

(b)

Interpretation Introduction

Interpretation:

To develop expressions for mole fractions of reacting species as function of reaction coordinate for

  $2H_{2}S(g)+3O_2(g)\to 2H_{2}O(g)+2S{O}_2(g)$

Concept Introduction :

The mole fractions y_i of the species present are related to $\epsilon$ by

  $y_i=\frac{n_i}{n}=\frac{n_{i_0}+v_i\epsilon }{n_0+v\epsilon }$

Answer to Problem 14.1P

  $y_i=\frac{n_i}{n}=\frac{n_{i_0}+v_i\epsilon }{n_0+v\epsilon }$

  $y{H}_{2}S=\frac{3-2\epsilon }{8-\epsilon }$

  $y{O}_2=\frac{5-3\epsilon }{8-\epsilon }$

  $y{H}_{2}O=\frac{2\epsilon }{8-\epsilon }$

  $yS{O}_2=\frac{2\epsilon }{8-\epsilon }$

Explanation of Solution

  $2H_{2}S(g)+3O_2(g)\to 2H_{2}O(g)+2S{O}_2(g)$

  $v_i$ = Moles of products − Moles of reactants

  $v=v_i=+2+2-2-3=-1$

  $n_0=3mol(H_{2}S)+5mol\left(O_2\right)=8$

The mole fractions y_i of the species present are related to $\epsilon$ by

  $y_i=\frac{n_i}{n}=\frac{n_{i_0}+v_i\epsilon }{n_0+v\epsilon }$

  $y{H}_{2}S=\frac{3-2\epsilon }{8-\epsilon }$

  $y{O}_2=\frac{5-3\epsilon }{8-\epsilon }$

  $y{H}_{2}O=\frac{2\epsilon }{8-\epsilon }$

  $yS{O}_2=\frac{2\epsilon }{8-\epsilon }$

(c)

Interpretation Introduction

Interpretation:

To develop expressions for mole fractions of reacting species as function of reaction coordinate for

  $6N{O}_2(g)+8N{H}_3(g)\to 7N_2(g)+12H_{2}O(g)$

Concept Introduction :

The mole fractions y_i of the species present are related to $\epsilon$ by

  $y_i=\frac{n_i}{n}=\frac{n_{i_0}+v_i\epsilon }{n_0+v\epsilon }$

Answer to Problem 14.1P

  $yN{O}_2=\frac{3-6\epsilon }{8+5\epsilon }$

  $yN{H}_3=\frac{4-8\epsilon }{8+5\epsilon }$

  $y{N}_2=\frac{1+7\epsilon }{8+5\epsilon }$

  $y{H}_{2}O=\frac{12\epsilon }{8+5\epsilon }$

Explanation of Solution

  $6N{O}_2(g)+8N{H}_3(g)\to 7N_2(g)+12H_{2}O(g)$

  $v_i$ = Moles of products − Moles of reactants

  $v=v_i=+7+12-6-8=5$

  $n_0=3molN{O}_2+4molN{H}_3+1mol{N}_2=8$

The mole fractions y_i of the species present are related to $\epsilon$ by

  $y_i=\frac{n_i}{n}=\frac{n_{i_0}+v_i\epsilon }{n_0+v\epsilon }$

  $yN{O}_2=\frac{3-6\epsilon }{8+5\epsilon }$

  $yN{H}_3=\frac{4-8\epsilon }{8+5\epsilon }$

  $y{N}_2=\frac{1+7\epsilon }{8+5\epsilon }$

  $y{H}_{2}O=\frac{12\epsilon }{8+5\epsilon }$