Question

At ${527}^{\circ }C$, the reaction given below has $K_c=4$
$N{H}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$
What is the $K_P$ for the reaction?
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$.

• A. $16\times (800R{)}^2$
• B. ${\left(\frac{800R}{4}\right)}^{-2}$
• C. ${\left(\frac{1}{4\times 800R}\right)}^2$
• D. None of these

Answer 1

The correct option is C ${\left(\frac{1}{4\times 800R}\right)}^2$

Given reaction is :-

${NH}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$

$△n=$ change in no. of moles of gaseous species

$=(\frac{3}{2}+\frac{1}{2})-1=2-1=1$

Given, $Kc=4$

Now, we have, $Kp=Kc{\left(RT\right)}^{△n}$

where, $R=$ universal gas constant

$T=$ temperature in $K=527+273=800K$

So, $Kp=4\times {\left(800R\right)}^1$

Now, for the reaction,

$N_2+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

$K^{{}^{\prime }}p={\left(\frac{1}{Kp}\right)}^2$

so, $K^{1}p={\left(\frac{1}{4\times 800R}\right)}^2$