Question

At $450\, K, K_p = 2.0 \times 10^{10}/bar $ for the given reaction at equilibrium. $2SO_2{(g)} + O_2\rightleftharpoons 2SO_3(g). $ What is $K_c $ at this temperature?

• A. $8 \times 10^{-2} M^{-1} $
• B. $5.47 \times 10^{-11} M^{-1} $
• C. $3 \times 10^{-5} M^{-1} $
• D. $7.47 \times 10^{11} M^{-1} $

Answer 1

Answer: (D)

$K_p =K_c(RT)\Delta n or K_c = \frac{K_p}{(RT)\Delta n} $
$\Delta n = 2 - (2 + 1) = -1 , T = 450 K, $
$R = 0.083 \,bar\, K^{-l} mol^{-1} $
$\therefore \frac{K_{p}}{\left(RT\right)^{-1}} =K_{P}\left(RT\right)$
$=2.0 \times 10^{10} bar ^{-1} \times 0.0831 \,L \,bar\, K ^{-1} mol ^{-1} \times 450 K$ $=74.8 \times 10^{10} \, mol ^{-1}=7.48 \times 10^{11}\, L \,mol ^{-1}$