Question

The value of ${\mathrm{K}}_{\mathrm{c}}$ is $64$ at $800K$ for the reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3$. The value of ${\mathrm{K}}_{\mathrm{c}}$ for the following reaction is :

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

• A. $\frac{1}{4}$
• B. $8$
• C. $\frac{1}{8}$
• D. $\frac{1}{64}$

Answer 1

The correct option is C

$\frac{1}{8}$

The explanation for the correct option:

C: $\frac{1}{8}$

Step 1: Given data

• For reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3$, Value of ${\mathrm{K}}_{\mathrm{c}}=64$

Step 2: Compute the value of ${\mathrm{K}}_{\mathrm{c}}$ for new reaction

• The value of ${\mathrm{K}}_{\mathrm{c}}$ for ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3$ will be,
• $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{c}}& =& {\left(\frac{1}{64}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& \frac{1}{8}\end{array}$

Final Answer: Hence, option (C) is the correct answer.