Answer to Question #346243 in General Chemistry for Ashter

Question

For a certain process at 127°C, ΔG = -16.20 kJ and ΔH = -17.0 kJ. What is the entropy change for this process at this temperature? Express your answer in the form, ΔS = ____________________ J/K.

Answer 1

∆G = -16,2 kJ;

∆H = -17 kJ;

T = 127°C = 400 K;

∆S = ?

Using the formula ∆G = ∆H - T * ∆S , we derive the formula for finding ∆S:

∆S = (∆H - ∆G) / T

∆S = ( (-17) - (-16,2) ) / 400

∆S = -0,002 kJ/K = -2 J/K

ANSWER: ∆S = -2 J/K