Solveeit Logo
Login

Question

Chemistry Question on Equilibrium Constant

At 1000 K1000\, K, the equilibrium constant. KCK_{C} for the reaction 2NOCl(g)⇌2NO(g)+Cl2(g)2 NOCl (g) \rightleftharpoons 2 NO (g)+ Cl _{2}(g) is 4.0×10−6mol L−14.0 \times 10^{-6} mol\, L ^{-1}. The KPK_{P} (in bar) at the same temperature is (R=0.083 L\left(R=0.083 \,L\right. bar K−1 mol−1)\left. K ^{-1}\, mol ^{-1}\right)

A

3.32×10−63.32 \times 10^{-6}

B

3.32×1043.32 \times 10^{4}

C

3.32×10−43.32 \times 10^{-4}

D

3.32×10−33.32 \times 10^{-3}

Answer

3.32×10−43.32 \times 10^{-4}

Explanation

Solution

Given, KC=4×10−6mol/LK_{C}=4 \times 10^{-6} mol / L
2NOCl(g)⇌2NO(g)+Cl2(g)2 NOCl (g) \rightleftharpoons 2 NO (g)+ Cl _{2}(g)
Δn=\Delta n= product mole −- reactant mole

=3−2=1=3-2=1

From KpK_{p} equation

Kp=KC×(RT)ΔnK_{p}=K_{C} \times(R T)^{\Delta n}
=4×10−6mol/L=4 \times 10^{-6} mol / L
×(0.0831 L bar  k−1 mol−1×1000 K)2\times\left(0.0831\, L \text { bar }\, k ^{-1}\, mol ^{-1}\right. \times 1000\, K )^{2}
Kp=3.32×10−4K_{p}=3.32 \times 10^{-4} bar