Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
0 votes

What is the Equilibrium constant for the formation of one mole of gaseous $HI$ from the $H_2(g)$ and $I_2(g)$?

Given: The rate of the elementary reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ at $25^\circ C$ is given by: Rate = $1.7 \times 10^{-18} [H_2] [I_2]$ . The rate of decomposition of gaseous HI to $H_2(g)$ and $I_2(g)$ at $25^\circ C$ is Rate = $2.4 \times 10^{-21} [HI]^2$.

(a) 708

(b) 354

(c) 0.0014

(d) 26.6

Can you answer this question?

1 Answer

0 votes
  • For a reversible reaction, at equilibrium, Rate of Forward reaction = Rate of Backward reaction
  • Rate of a reaction = Equilibrium Constant $\times$ Product of Concentration of reactants or products
Answer: 26.6
For one mole of gaseous HI, $\frac{1}{2} H_2 (g) + \frac{1}{2} I_2 (g) \rightleftharpoons HI (g)$
$K_c = \frac{[HI]}{[H_2]^{\frac{1}{2}}[I_2]^{\frac{1}{2}}}$
For the reaction given in the question,
$H_2 (g) + I_2 (g) \rightleftharpoons 2HI (g)$
Rate of Forward reaction, $R_f = 1.7 \times 10^{-18} [H_2] [I_2]$
Rate of Backward reaction, $R_b = 2.4 \times 10^{-21} [HI]^2$
At equilibrium,
Rate of Forward reaction = Rate of Backward reaction
$\therefore R_f = R_b$
$\Rightarrow 1.7 \times 10^{-1 8} [H_2][I_2] = 2.4 \times 10^{-21} [HI]^2$
$\Rightarrow \frac{1.7 \times 10^{-18}}{2.4 \times 10^{-21}} = \frac{[HI]^2}{[H_2][I_2]}$
$\Rightarrow \frac{170}{24} \times 10^2 = K_c^2$
$\Rightarrow K_c = \sqrt{\frac{170}{24}} \times 10 = \sqrt{7.08} \times 10 = 2.66 \times 10 = 26.6$
answered Nov 23, 2013 by mosymeow_1
edited Mar 21, 2014 by mosymeow_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App