Question

For the reversible reaction, $N_2\left(g\right)+3H_2\left(g\right)=2N{H}_3\left(g\right)$ at ${500}^o$C, the value of $K_p$ is $1.44\times {10}^{-5}$ when partial pressure is measured in the atmosphere. The corresponding value of $K_c$, with concentration in $mollitr{e}^{-1}$ is:

• A. $1.44\times {10}^5/(0.082\times 500{)}^{-2}$
• B. $1.44\times {10}^{-5}/(8.314\times 773{)}^{-2}$
• C. $1.44\times {10}^{-5}/(0.082\times 773{)}^2$
• D. $1.44\times {10}^{-5}/(0.082\times 773{)}^{-2}$

Answer 1

The correct option is A $1.44\times {10}^5/(0.082\times 500{)}^{-2}$

$N_2\left(g\right)+3H_2\left(g\right)=2N{H}_3\left(g\right)$

$K_c=\frac{K_p}{(RT{)}^{\Delta n}}$

$K_c=\frac{1.44\times {10}^{-5}}{(0.082\times 500{)}^{-2}}$ ($\Delta n=-2$)