The standard enthalpy and standard entropy of decomposition of N₂ O₄ to NO₂ are 55.0 kJ mol⁻¹ and 175.0 J / K / mol respectively. The standard free energy change for this reaction at 25° C in J mol⁻¹ is ________ (Nearest integer)

Solution

Chemistry ma’am · 3 steps

Put Δ H in joules and T in kelvin, then Δ G = Δ H - TΔ S does the rest.

1. Match the units. Δ H° is given in kJ and Δ S° in J. The answer is asked in J, so write Δ H° = 55.0 kJ mol⁻¹ = 55000 J mol⁻¹.
2. Temperature in kelvin. 25°C is taken as T = 298 K. This is the value the answer key uses.

   Δ G° = Δ H° - TΔ S° = 55000 - 298 × 175 = 55000 - 52150

3. So. Δ G° = 2850 J mol⁻¹. It is small and positive, so the decomposition is slightly non-spontaneous under standard conditions at 298 K. The answer is 2850.

The common slip: -52095. Leaving Δ H° as 55 while Δ S° stays in joules gives 55 - 52150 = -52095, which is far too large and has the wrong sign.

The idea to keep. Before you use Δ G = Δ H - TΔ S, check the units. Δ H usually comes in kJ and Δ S in J, so convert one of them, and always put T in kelvin.

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