Question

The reaction below has an equilibrium constant K p =2.2× 10 6 at 298 K. 2 COF 2 (g)⇌ CO 2 (g)+ CF 4 (g) Part A Calculate K p for the reaction below. COF 2 (g)⇌ 1 2 CO 2 (g)+ 1 2 CF 4 (g) Part B Calculate K p for the reaction below. 2 3 COF 2 (g)⇌ 1 3 CO 2 (g)+ 1 3 CF 4 (g) Part C Calculate K p for the reaction below. 2 CO 2 (g)+2 CF 4 (g)⇌4 COF 2 (g)

Answer 1

Given that   2 COF2 (g)⇌ CO2 (g)+ CF4 (g) -- Eq(1)      Kp = 2.2 × 10⁶

A) COF₂ (g)⇌ 1/2 CO2 (g)+ 1/2 CF4 (g)

     On multiplying Eq(1) with 1/2 , we will get this reaction.

Hence,

    Kp for this reaction = [Kp]^1/2 = [2.2× 10⁶]^1/2 = 1483.2

B)   2/3 COF2 (g)⇌ 1/3 CO 2 (g)+ 1/3 CF4

    On multiplying Eq(1) with 1/3 , we will get this reaction.

Hence,

    Kp for this reaction = [Kp]^1/3 = [2.2× 10⁶]^1/3 = 130.06

C)    2CO2 (g)+ 2 CF4 (g)⇌ 4COF2 (g)

    This reaction is obtained on reversing and multiplying the Eq(1) with 2.

   Hence,

   Kp for this reaction = 1/(Kp)² = 1/(2.2× 10⁶)² = 2.06 x 10^-13