At 298 K, the equilibrium constant is 2× 10¹⁵ for the reaction :
Cu(s)+2Ag⁺(aq)⇌ Cu²⁺(aq)+2Ag(s)
The equilibrium constant for the reaction
1/2Cu²⁺(aq)+Ag(s)⇌ 1/2Cu(s)+Ag⁺(aq)
is x× 10⁻⁸. The value of x is _____________. (Nearest Integer)

Solution

Chemistry ma’am · 4 steps

The new reaction is the old one reversed and halved, so invert K and then take its square root.

1. Compare the reactions. The second reaction runs the first one backwards, so its constant is 1/K. Every coefficient is also multiplied by 1/2, so that constant is raised to the power 1/2.

   K' = (1/K)^(1/2) = 1/√(2 × 10¹⁵)

2. Make the power even. Write 2 × 10¹⁵ = 20 × 10¹⁴ so the square root is clean: √(20 × 10¹⁴) = 4.472 × 10⁷.
3. Invert. K' = 1/(4.472 × 10⁷) = 0.2236 × 10⁻⁷ = 2.236 × 10⁻⁸.
4. So. x = 2.236, which rounds to 2.

The idea to keep. Reversing a reaction inverts K. Multiplying all coefficients by n raises K to the power n. Do both when both happen, and rewrite the power of ten as even before a square root.

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