The value of KC is 64 at 800 K for the reaction

Question:

The value of $K_{C}$ is 64 at $800 K$ for the reaction

$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$

The value of $K_{C}$ for the following reaction is :

$\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})$

  1. $\frac{1}{4}$

  2. $\frac{1}{8}$

  3. 8

  4. $\frac{1}{64}$


Correct Option: , 2

Solution:

$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \rightarrow \mathrm{K}_{\mathrm{C}}=64$

$2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow \mathrm{K}_{\mathrm{C}}=\frac{1}{64}$

$\mathrm{NH}_{3} \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightarrow \mathrm{K}_{\mathrm{C}}=\left(\frac{1}{64}\right)^{1 / 2}=\frac{1}{8}$

 

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