JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

The value of K C is 64 at 800 K for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
The value of K C for the following reaction is :
NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)

  1. (a) 8
  2. (b) 1/8
  3. (c) 1/4
  4. (d) 1/64
Answer and solution

Answer (b) 1/8

Solution

Chemistry ma’am · 3 steps

This reaction is the original one reversed and then halved, so flip K_C first and then take its square root.

  1. Reverse it. Writing a reaction backwards turns K_C into its reciprocal. For 2NH₃ ⇌ N₂ + 3H₂, the constant is 1/64.
  2. Halve it. Multiplying every coefficient by 1/2 raises K_C to the power 1/2. Each concentration term in the expression gets half its old power, so the whole constant becomes a square root.

    K_C' = (1/64)^(1/2) = 1/8

  3. So. For NH₃ ⇌ 1/2 N₂ + 3/2 H₂, K_C = 1/8, which is option (b).

The idea to keep. Reversing a reaction inverts K. Multiplying the coefficients by n raises K to the power n. Apply both changes and you can handle any rewritten equation.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Chemical Equilibrium

Asked about 90 times in JEE Main since 2020. This question is from 2020.