Question

The reaction between $N_2$ and $H_2$ to form ammonia has $K_C=6\times {10}^{-2}$ at the temperature ${500}^{o}C$. The numerical value of $K_p$ for this reaction is;

• A. $1.5\times {10}^{-5}$
• B. $1.5\times {10}^5$
• C. $1.5\times {10}^{-6}$
• D. $1.5\times {10}^6$

Answer 1

The correct option is A $1.5\times {10}^{-5}$

$N_2+3H_2\rightleftharpoons 2N{H}_3$

Given : $K_C=6.0\times {10}^{-2}:T={500}^{\circ }C=500+273=773K$

$\Delta n=2-3-1=-2$

$\therefore K_p=K_c(RT{)}^{\Delta n}=6\times {10}^{-2}\times (0.0821\times 773{)}^{-2}$

$=6\times {10}^{-2}\times (63.46{)}^{-2}=\frac{6\times {10}^{-2}}{(63.46{)}^2}=\frac{6\times {10}^{-2}}{4027.17}=0.00148\times {10}^{-2}$

$K_p=1.5\times {10}^{-2}\times {10}^{-3}=1.5\times {10}^{-5}$