Question

Two moles of h2 and three moles of i2 are taken in 2 dm3 vessel and heated. if the equilibrium mixture contains 0.8 moles of hi, calculate kp and kc for the reaction

Answer 1

Two moles of H2 and three moles of I2 are taken in 2dm3 vessel and heat.if the equilibrium mixture contain 0.8 moles of HI, calculate Kp and Kc for the reaction H 2(g] + I 2(g] ⇌ 2 HI ...

Answer 2

Answer: $K_c$ = 0.15

Explanation:

$H_2+I_2\rightleftharpoons 2HI$

initially  2        3                 0

at eq'm   2-x     3-x           2x     

Given : moles of $HI$ at equilibrium= 2x = 0.8, x=0.4

Thus moles of hydrogen at equilibrium = (2-x) = (2-0.4) = 1.6

Thus moles of iodine at equilibrium = (3-x) = (3-0.4) = 2.6

So eq'm constant will be:

$K_c=\frac{[HI]^2}{[H_2]\times[I_2]}$

$concentration=\frac{\text{moles}}{\text{Volume in Litres}}$

${\text {concentration of hydrogen iodide}}=\frac{0.8}{2L}=0.4M$

${\text {concentration of hydrogen}}=\frac{1.6}{2L}=0.8M$

${\text {concentration of iodine}}=\frac{2.6}{2L}=1.3M$

$K_c=\frac{[0.4]}^2{[0.8]\times [1.3]}$

$K_c=0.15$

$K_p=K_c(RT)^\Delta {n_g}$

R= gas constant

T = temperature

$\Delta n_g$= number of gaseous moles of products- number of gaseous moles of reactants