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Question

Chemistry Question on Equilibrium

For the reaction 2NO2(g)<=>2NO(g)+O2(g) {2NO2(g) <=> 2\,NO(g) + O2(g)} (Kc=1.8×106at184C)\left(K_{c} = 1.8\times10^{-6} \,at \,184^{\circ}C\right) (R=0.00831kJ/(mol.K))\left(R = 0.00831 \,k J/\left(mol. K\right)\right) When KpK_{p} and KcK_{c} are compared at 184184^{\circ} C it is found that :

A

whether KpK_p is greater than, less than or equal to KcK_c depends upon the total gas pressure

B

Kp=KcK_p = K_c

C

KpK_p is less than KcK_c

D

KpK_p is greater than KcK_c

Answer

KpK_p is greater than KcK_c

Explanation

Solution

2NO2(g)<=>2NO(g)+O2(g) {2NO2(g) <=> 2\,NO(g) + O2(g)} Kc=1.8×106at184C(=457K)K_{c} = 1.8\times10^{-6} \,at \,184^{\circ}C ( = 457\, K) R=0.00831kJmol1K1R = 0.00831 \,k J \,mol^{-1} K^{-1} Kp=Kc(RT)Δng K_{p}=K_{c}\left(RT\right)^{\Delta n_{g}} where, Δng=\Delta n_{g} = (gaseous products - gaseous reactants) =32=1=3-2=1 Kp=1.8×106×0.00831×457\therefore\quad K_{p} = 1.8 \times 10^{-6 }\times 0.00831 \times 457 =6.836×106>1.8×106= 6.836 \times10^{-6} > 1.8 \times10^{-6} Thus Kp>KcK_{p} > K_c