Assuming ideal behaviour, the magnitude of log K for the following reaction at 25°C is x × 10⁻¹. The value of x is ______________. (Integer answer)
3HC ≡ CH_((g)) ⇌ C₆H₆₍ₗ₎
[Given : Δ _fGᵒ(HC ≡ CH) = - 2.04 × 10⁵ J mol⁻¹ ; Δ _fGᵒ(C₆H₆) = - 1.24 × 10⁵ J mol⁻¹ ; R = 8.314 J K⁻¹ mol⁻¹]

Solution

Chemistry ma’am · 3 steps

Build Δᵣ G° as products minus reactants, keeping the 3 on acetylene, then turn it into log K.

1. Find ΔrG°. Take products minus reactants, and multiply each Δ_f G° by its coefficient: Δᵣ G° = (-1.24 × 10⁵) - 3(-2.04 × 10⁵) = -1.24 × 10⁵ + 6.12 × 10⁵ = +4.88 × 10⁵ J mol⁻¹.

   log K = -(Δᵣ G°)/(2.303 RT)

2. Put in numbers. At 25°C, T = 298 K, so 2.303 × 8.314 × 298 = 5705.8 J mol⁻¹. Then log K = -(4.88 × 10⁵)/5705.8 = -85.5.
3. So. The magnitude of log K is 85.5 = 855 × 10⁻¹, so x = 855.

The common slip: 140. Leaving out the coefficient 3 on acetylene gives Δᵣ G° = 8.0 × 10⁴ J mol⁻¹, so log K ≈ -14.0 and x comes out as 140.

The idea to keep. Always weight each Δ_f G° by its stoichiometric coefficient before using Δ G° = -2.303RT log K. A positive Δ G° means a negative log K.

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