Chemical Equilibrium

Given reaction:$\mathrm{A_2O_4}(g) \rightleftharpoons 2\mathrm{AO_2}(g)$At 298 K , the equilibrium constant in concentration units is$K_c = x\ \mathrm{mol}\ \mathrm{L}^{-1}$.We need to find the approximate value of$K_P$.Relation between $K_P$ and$K_C$For a gaseous reaction:$K_P = K_C (RT)^{\Delta n}$where$\Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}$Step 1: Determine$\Delta n$Moles of gaseous products$=2$Moles of gaseous reactants$=1$$\Rightarrow \Delta n = 2-1 = +1$Step 2: Substitute into the relation$K_P = K_C (RT)^1 = K_C (RT)$Step 3: Plug in the given valuesAt$T = 298\ \mathrm{K}$,$R = 0.082\ \mathrm{L}\ \mathrm{atm}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}$$RT = 0.082 \times 298 = 24.4\ \mathrm{L}\ \mathrm{atm}\ \mathrm{mol}^{-1}$Step 4: Therefore,$\;K_P = K_C \times 24.4$$\;K_P = 24.4 x$Final Answer: Option A -$24.4 x$