Question

The heat of reaction is -85.0 kj and the entropy change for the reaction is 9.50 j/k. at 298 k, what is the value of delta g, and will the process be spontaneous or non-spontaneous?

A. -82.2 kj; spontaneous
B. -82.2 kj; non-spontaneous
C. -87.8 kj; spontaneous
D. -2,746 kj; non-spontaneous

Answer 1

The value of delta G at 298 K is calculated to be approximately -87.8 kJ. Since this value is negative, the reaction is classified as spontaneous. The correct option is C. -87.8 kJ; spontaneous.

Step-by-step explanation:

The value of delta G for a reaction can be calculated using the formula: Delta G = Delta H - T*Delta S.

Where Delta H is the enthalpy change, T is the temperature in kelvins, and Delta S is the entropy change. Given that the heat of the reaction (Delta H) is -85.0 kJ and the entropy change (Delta S) for the reaction is 9.50 J/K, we can calculate delta G at 298 K as follows:

Delta G = (-85.0 kJ) - (298 K) * (9.50 J/K * 1 kJ/1000 J)

Delta G = (-85.0 kJ) - (298 * 0.0095 kJ)

Delta G = -85.0 kJ - 2.831 kJ

Delta G = -87.831 kJ or approximately -87.8 kJ.

Since Delta G is negative, the process is spontaneous at 298 K. The correct option is C. -87.8 kJ; spontaneous, and with this computed value, it indicates that the reaction would proceed without external influence.