For the reaction, 2NO₂(g) ⇌ N₂O₄(g), when ΔS = -176.0 JK⁻¹ and ΔH = -57.8 kJ mol⁻¹, the magnitude of ΔG at 298 K for the reaction is ___________ kJ mol⁻¹. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Turn Δ S into kJ first, then let the two minus signs in -TΔ S do their work.

1. Δ G = Δ H - TΔ S

2. Match the units. Δ H is in kJ but Δ S is in J, so convert: Δ S = -176.0 J K⁻¹ = -0.176 kJ K⁻¹.
3. Find TΔS. TΔ S = 298 × (-0.176) = -52.448 kJ mol⁻¹. Subtracting a negative number means adding, so -TΔ S = +52.448 kJ mol⁻¹.
4. Put in numbers. Δ G = -57.8 + 52.448 = -5.352 kJ mol⁻¹. It is still negative, so the reaction is just barely spontaneous at 298 K.
5. So. The magnitude of Δ G is 5.352 kJ mol⁻¹, which rounds to 5.

The common slip: 110. Treating -TΔ S as -52.448 instead of +52.448 gives -110.248, a magnitude of 110 rather than 5.

The idea to keep. When Δ H and Δ S are both negative, the -TΔ S term works against Δ H. Always bring both terms to the same unit before you subtract.

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