Consider the equilibrium, H 2+ I 2 leftharpoons 2 HI Calculate the equilibrium constant of the reverse reaction when the equilibrium concentration of H 2, I 2 and HI are 1.14 × 10-2, 0.12 × 10-2 and 2.52 × 10-2 mol L -1 respectively

Question

Consider the equilibrium, H 2+ I 2 leftharpoons 2 HI Calculate the equilibrium constant of the reverse reaction when the equilibrium concentration of H 2, I 2 and HI are 1.14 × 10-2, 0.12 × 10-2 and 2.52 × 10-2 mol L -1 respectively (A) 46.4 (B) 0.021 (C) 18.42 (D) 0.054

Tardigrade Admin · Accepted Answer

H 2+ I 2 leftharpoons 2 HI Reverse of above reaction is 2 HI leftharpoons H 2+ I 2 Equilibrium constant(K)=([ H 2][ I 2]/[ HI ]2) Given, Concentration of H 2=1.14 × 10-2 mol / L Concentration of I 2=0.12 × 10-2 mol / L Concentration of HI =2.52 × 10-2 mol / L K=([1.14 × 10-2][0.12 × 10-2]/[252 × 10-2]2) K=0.021

Q. Consider the equilibrium, H2​+I2​⇌2HI Calculate the equilibrium constant of the reverse reaction when the equilibrium concentration of H2​,I2​ and HI are 1.14×10−2,0.12×10−2 and 2.52×10−2molL−1 respectively

46.4

0.021

18.42

0.054

H2​+I2​⇌2HI Reverse of above reaction is 2HI⇌H2​+I2​ Equilibrium constant(K)=[HI]2[H2​][I2​]​ Given, Concentration of H2​=1.14×10−2mol/L Concentration of I2​=0.12×10−2mol/L Concentration of HI=2.52×10−2mol/L K=[252×10−2]2[1.14×10−2][0.12×10−2]​ K=0.021

Q. Consider the equilibrium, $H_{2} + I_{2} ⇌ 2 H I$
Calculate the equilibrium constant of the reverse reaction when the equilibrium concentration of $H_{2}, I_{2}$ and $H I$ are
$1.14 \times 1 0^{- 2}, 0.12 \times 1 0^{- 2}$ and $2.52 \times 1 0^{- 2} m o l L^{- 1}$
respectively