Question

Assume that decomposition of $HN{O}_3$ can be represented by the following equation
$4HN{O}_3\left(g\right)\rightleftharpoons 4N{O}_2\left(g\right)+2H_{2}O\left(g\right)+O_2\left(g\right)$
and the reaction approaches equilibrium at $400K$ temperature and $30atm$ pressure. At equilibrium partial pressure of $HN{O}_3$ is $2atm$.
Calculate $K_c$ in ${\left(mol/L\right)}^3$ at $400K$ :(Use : $R=0.08atm-L/mol-K$)

• A. $4$
• B. $8$
• C. $16$
• D. $22.8$

Answer 1

The correct option is D $22.8$

Let initial mole of $HN{O}_3$ be $1$ Total number of moles= $1+3x$; Total Pressure= $P_T=30atm$

$4HN{O}_3\rightleftharpoons 4N{O}_2+2H_{2}O+O_2$

at $eq{b}^m$ $1-4x$ $4x$ $2x$ $x$

mole fraction $\frac{1-4x}{1+3x}$ $\frac{4x}{1+3x}$ $\frac{2x}{1+3x}$ $\frac{x}{1+3x}$

Partial Pressure$\left[\frac{1-4x}{1+3x}\right]\times P_T$ $\left[\frac{4x}{1+3x}\right]P_T$ $\left[\frac{2x}{1+3x}\right]P_T$ $\left[\frac{x}{1+3x}\right]P_T$

As given partial pressure of $HN{O}_3=2atm$

$\Rightarrow \left[\frac{1-4x}{1+3x}\right]\times 30=2atm$

On solving we get $x=0.22$

$K_P=\frac{\left[X_{N{O}_2}{]}^4\right[X_{H_{2}O}{]}^2\left[X_{O_2}\right]}{[X_{HN{O}_3}{]}^4}[P{]}^{\Delta ng}$

$\Rightarrow \Delta ng=7-4=3$

$K_P=\frac{\left[4x{]}^4\right[2x{]}^2\left[x\right]}{[1-4x{]}^2}{\left[\frac{30}{1+3x}\right]}^3$

On putting the value of $x$, we have

$K_P=7.27\times {10}^5$

As $K_P=K_C(RT{)}^{\Delta ng}$

$\Rightarrow K_C=\frac{K_P}{(RT{)}^{\Delta ng}}=\frac{7.27\times {10}^5}{(0.08\times 400{)}^3}=22.18$

$\Rightarrow K_C=22.18$