Consider the following reaction approaching equilibrium at 27°C and 1 atm pressure
A+B ⇌(k_f=10³) C+D
The standard Gibb's energy change (ΔᵣG^θ) at 27°C is (-) ___________ kJ mol⁻¹ (Nearest integer).
(Given : R=8.3 J K⁻¹ mol⁻¹ and ln 10=2.3)

Solution

Chemistry ma’am · 3 steps

At equilibrium the forward and reverse rates are equal, so K is just k_f divided by kᵣ.

1. Find K. At equilibrium, rate forward equals rate backward: k_f[A][B] = kᵣ[C][D]. Rearranging gives K = k_f/kᵣ = 10³/10² = 10.

   Δᵣ G° = -RT ln K

2. Put in numbers. Use T = 27 + 273 = 300 K and ln 10 = 2.3. Then Δᵣ G° = -8.3 × 300 × 2.3 = -5727 J mol⁻¹.
3. So. Δᵣ G° = -5.727 kJ mol⁻¹, which rounds to -6. The blank takes 6.

The common slip: 1. Putting T = 27 instead of 300 K gives -0.515 kJ mol⁻¹, which rounds to 1, not 6. Always convert to kelvin first.

The idea to keep. Whenever both rate constants are given, K = k_f/kᵣ. Feed that K into Δᵣ G° = -RT ln K, with T in kelvin and the answer turned into kJ at the end.

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