At 450 K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.

Question:

At $450 \mathrm{~K}, K_{\mathrm{p}}=2.0 \times 10^{10} /$ bar for the given reaction at equilibrium.

$2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \longleftrightarrow 2 \mathrm{SO}_{3(g)}$

What is Kat this temperature?

Solution:

For the given reaction,

$\Delta n=2-3=-1$

$T=450 \mathrm{~K}$

$R=0.0831$ bar $\mathrm{L}$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$

$K_{\mathrm{p}}=2.0 \times 10^{10} \mathrm{bar}-1$

We know that,

$K_{\mathrm{P}}=K_{\mathrm{C}}(R T) \Delta n$

$\Rightarrow 2.0 \times 10^{10} \mathrm{bar}^{-1}=K_{C}\left(0.0831 \mathrm{~L} \text { bar } \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 450 \mathrm{~K}\right)^{-1}$

$\Rightarrow K_{\mathrm{C}}=\frac{2.0 \times 10^{10} \mathrm{bar}^{-1}}{\left(0.0831 \mathrm{~L} \text { bar } \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 450 \mathrm{~K}\right)^{-1}}$

$=\left(2.0 \times 10^{10} \mathrm{bar}^{-1}\right)\left(0.0831 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 450 \mathrm{~K}\right)$

$=74.79 \times 10^{10} \mathrm{~L} \mathrm{~mol}^{-1}$

$=7.48 \times 10^{11} \mathrm{~L} \mathrm{~mol}^{-1}$

$=7.48 \times 10^{11} \mathrm{M}^{-1}$