The equilibrium constant K_p for the following reaction at 191 ,^… - Vidyadip

MCQ

The equilibrium constant $K_p$ for the following reaction at $191\,^oC$ is $1.24$. What is the value of $K_c$ ? $B\left( s \right) + \frac{3}{2}{F_2}_{\left( g \right)} \rightleftharpoons B{F_3}_{\left( g \right)}$

A
$6.7$
B
$0.61$
C
$8.3$
✓
$7.6$

✓

Answer

Correct option: D.

$7.6$

d
$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}=(\mathrm{RT})^{-1 / 2}=\frac{\mathrm{K}_{\mathrm{e}}}{(\mathrm{RT})^{1 / 2}}$

$\Rightarrow \mathrm{K}_{\mathrm{e}}=1.24 \times(0.0821 \times 464)^{1 / 2}=7.65$

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