Question

If $K_p$ for the given reaction is $1.44\times {10}^{-5}units$, then the value of $K_c$ will be:

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

Answer 1

Answer: (D)

$\frac{1.44\times {10}^{-5}}{{(0.082\times 773)}^{-2}}mo{l}^{-2}{L}^2$

The given reaction is:-

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right);{△H}^0=-22.4KJ/mol$

$T={500}^{0}C=500+273=773K$

$R=0.0821Latm{K}^{-1}{mol}^{-1}$

Now, $△n=2-1-3=-2$

So, we have, $K_P=K_C{\left(RT\right)}^{△n}$

$\Rightarrow K_C=\frac{K_P}{{\left(RT\right)}^{△n}}$

$=\frac{1.44\times {10}^{-5}}{{(0.0821\times 773)}^{-2}}{mol}^{-2}{L}^2$

Hence, the correct option is $\text{D}$