Reversible Reaction

Chemistry Level 2

In reversible reaction $Δn=\dfrac{1}{2}$ .

So In the reaction at which kelvin temperature

$K_{p}=40.5$

and $K_{c}=5.5$ ?

Submit your answer to two decimal places.

1 solution

Md Zuhair
Oct 28, 2017

We know the relation between $K_{p}$ and $K_{c}$ , it is

$\large{K_{p}=K_{c} \times (RT)^{\Delta n_{g}}}$

$\implies \large{40.5=5.5 \times (RT)^{\dfrac{1}{2}}}$

$\implies \large{\dfrac{40.5}{5.5} = (RT)^{\dfrac{1}{2}}}$

$\implies \large{(\dfrac{40.5}{5.5})^2 = (RT)}$

Now putting $R=0.0821 \space \space \text{L atm} \space \text{mol}^{-1} \space K^{-1}$

We get $\boxed{T= 661.26 K}$