For the reaction A (g) → B (g) the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of ΔᵣG for the reaction at 300 K and 1 atm in J mol⁻¹ is – xR, where x is _______. (Rounded off to the nearest integer)
[R = 8.31 J mol –1 K⁻¹ and ln 10 = 2.3)

Solution

Chemistry ma’am · 4 steps

At 1 atm you are in the standard state, so Δᵣ G° = -RT ln K and you can leave R as a symbol.

1. Pick the relation. At 1 atm every gas is in its standard state, so the Δᵣ G asked for is Δᵣ G°. It links to the equilibrium constant through one equation.

   Δᵣ G° = -RT ln K

2. Find ln K. ln 100 = ln 10² = 2 ln 10 = 2 × 2.3 = 4.6.
3. Put in numbers. Δᵣ G° = -R × 300 × 4.6 = -1380R J mol⁻¹. Do not substitute R = 8.31, because the answer is asked as a multiple of R.
4. So. Comparing with -xR gives x = 1380.

The common slip: 600. Using log₁₀ 100 = 2 in place of ln 100 = 4.6 gives 600, not 1380.

The idea to keep. When the answer is asked as a multiple of R, keep R as a symbol. And always use the natural log in Δ G° = -RT ln K, or multiply log₁₀ K by 2.303.

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