High School

For a given reaction, ΔS = +69.0 J/mol∙K, and the reaction is spontaneous at temperatures above the crossover temperature, 439 K. The value of ΔH = ________ kJ/mol, assuming that ΔH and ΔS do not vary with temperature.

A. 30.3
B. -1.57 x 10⁻⁴
C. 30.3
D. 6.36 x 10⁻⁴
E. 1.57 x 10⁻⁴

Answer :

Final answer:

The value of ΔH for the given reaction is -30.3 kJ/mol.

Explanation:

To find the value of ΔH for the given reaction, we need to use the equation: ΔG = ΔH - TΔS. Since the reaction is spontaneous, ΔG is negative. We are given that ΔS = +69.0 J/mol∙K and the reaction is spontaneous above the crossover temperature, 439 K. To calculate ΔH, we can rearrange the equation to solve for ΔH: ΔH = ΔG + TΔS. Plugging in the values, we get ΔH = -69.0 J/mol∙K * 439 K = -30,291 J/mol = -30.3 kJ/mol.

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