Question

For a given reaction, AH = -19.9 kJ/mol and AS = -55.5 J/K/mol. The reaction will have AG = 0 at K. Assume that AH and AS do not vary with temperature. A) 359 B) 2789 C) 298 D) 2.79 E) 0.359

Answer 1

The correct answer is option A.

Step-by-step explanation:

The expression of Gibbs's fee energy is given as:

`\Delta G=\Delta H-T\Delta S`

`\Delta G` = Change in Gibbs free energy

`\Delta S` = Change in an entropy

`\Delta H` = Enthalpy of reaction

T = Temperature at which reaction is going on.

We have:

`\Delta G=0 J/mol`

`\Delta S=-55.5 J/K mol`

`\Delta H=-19.9 kJ/mol=-19,900 J/mol`

T =?

`0 J/mol=-19,900 J/mol-(T* (-55.5 J/K mol))`

T = 359 K