logo

Ask Questions, Get Answers

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

The equilibrium constant $K_c$ for the reaction of $H_2$ with $I_2$ is 57.0 at 700 K $ H_2(g) + I_2(g) \rightleftharpoons 2HI $; $K_c$ = 57 at 700K. select correct statement:

(a) rate constant $k_f$ for the formation of HI is smaller than that of rate constant $k_b$ for the dissociation of HI

(b) $k_f > k_b$

(c) addition of catalyst increases value of $K_c$.

(d) addition of catalyst decreases value of $K_c$.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Rate of reaction = Equilibrium Constant $\times$ Product of concentration of reactants / products in the reaction
  • $R_x = k_x [A] [B]$ where $R_x$ = rate of reaction, $k_x$ = equilibrium constant and, A and B are reactants or products
  • At equilibrium, Rate of Forward reaction = Rate of backward reaction. $R_f = R_b.$
 
Answer: $k_f > k_b$
 
For forward reaction, $R_f = k_f [H_2][I_2]$
 
For backward reaction, $R_b = k_b [HI]^2$
 
At equilibrium, $R_f = R_b$
$\Rightarrow k_f [H_2] [I_2] = k_b [HI]^2 $
$\Rightarrow \frac{k_f}{k_b} = \frac{[HI]^2}{[H_2][I_2]} = K_c$ = 57
 
As, $K_c$ = 57
$k_f = 57 \times k_b$
$\therefore k_f > k_b$

 

answered Nov 23, 2013 by mosymeow_1
 

Related questions

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