Question

The Keq for the equilibrium below is 5.4 × 1013 at 480.0°C. 2NO (g) + O2 (g) 2NO2 (g) What is the value of Keq at this temperature for the following reaction? 4NO (g) + 2O2 (g) 4NO2 (g)

A) 5.4 × 1013 B) 5.4 × 10-13 C) 1.9 × 1012 D) 1.9 × 10-12 E) 2.9 × 1027

Answer 1

Sol:- '

Equilibrium constant i.e. K_eq is " the ratio of product of the molar concentration of products to the product of the molar concentration of the reactants, in which stoichiometric coefficient with respect to each reagent is raised to the power of concentration at equilibrium stage of the reaction".

Given first chemical equation is :-

2NO (g) + O₂ (g) <---------------> 2NO₂ (g) , K_eq = 5.4 × 10¹³ ..............(1)

Expression of K_eq for this equation is:

K_eq = [ NO₂ (g) ]² / [ NO (g)]² [ O₂ (g)] .................(2)

Multiply equation (1) by 2 on both the sides, we have

4 NO (g) + 2O₂ (g) <----------> 4 NO₂ (g) (which is the second equation)

Now expression of equilibrium constant say K_eq^' is :

K_eq^' = [ NO2 (g)]⁴ / [ NO (g]⁴ [ O₂ (g)]² ..................(3)

Comparing equations (2) and (3), we have the relation between K_eq and K_eq^' is :

K_eq^' = ( K_eq)²

so

K_eq^' = ( 5.4 × 10¹³ )²

K_eq^' = 2.9 x 10²⁷

Hence equilibrium constant for the second equation is 2.9 x 10²⁷ .