For a certain reaction at 300 K, K=10, then Δ G° for the same reaction is - ____________ × 10⁻¹ kJ mol⁻¹.
(Given R=8.314 JK⁻¹ mol⁻¹ )

Solution

Chemistry ma’am · 4 steps

Link Δ G° to K with -RT ln K, then read the answer in tenths of a kJ.

1. The link. At equilibrium, the standard Gibbs energy change is tied to the equilibrium constant. With base-10 logs, the factor 2.303 converts ln to log.

   Δ G° = -RT ln K = -2.303 RT log K

2. Put in numbers. With K = 10, log 10 = 1. So Δ G° = -2.303 × 8.314 × 300 × 1 = -5744 J mol⁻¹.
3. Change the unit. -5744 J mol⁻¹ = -5.744 kJ mol⁻¹ = -57.44 × 10⁻¹ kJ mol⁻¹.
4. So. The blank after the minus sign is 57.

The common slip: 25. Using RT log K without the 2.303 factor gives 2494 J mol⁻¹, which is about 25 in these units, not 57.

The idea to keep. Remember Δ G° = -2.303 RT log K. A K above 1 always gives a negative Δ G°. Always check the unit the blank asks for before you round.

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