Question

$K_p$ for a reaction at ${25}^{o}C$ in 10 atm is 8.568. Then $K_c$ for the reaction at ${25}^{o}C$ will be:
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

• A. $4.33\times {10}^{-1}M$
• B. $3.33\times {10}^{-2}M$
• C. $3.89\times {10}^{-1}M$
• D. none of these

Answer 1

The correct option is D none of these

We know, relation between the reactions constants as:
$K_p=K_c(RT{)}^{△n},$
As per reaction:
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ $△n=-2$
$\therefore 8.568\times (0.0821\times 313{)}^2=K_c$
$K_c\Rightarrow 5657.88$