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Consider a reaction with ΔH = 15 kJ and ΔS = 50 J · K−1 . Is the reaction spontaneous (a) at 10°C, (b) at 80°C?

Short Answer

Expert verified

When ΔH = 15 kJ and ΔS = 50 J · K−1, the reaction is spontaneous at 10°C, and at 80°C.

Step by step solution

01

 Definition and Formula

Enthalpy reflects the capacity to do a non-mechanical work and the capacity to release heat. It is denoted by H and its unit is joules.

Entropy indicates the degree of randomness of a system and is abbreviated as S. Its unit is J . K-1.

Free energy indicates the part of the total energy of a system which is available for useful work. A change in the free energy of a process is ΔG.

ΔG=ΔH−TΔS

For a spontaneous press process ΔG < zero and such processes are said to be exergonic. Non spontaneous process has a positive ΔG value and are said to be endergonic.

02

Solution

According to the question ΔH = 15 kJ and ΔS = 50 J · K−1

a) at 10°C

ΔG=ΔH−TΔSΔG=15−(10×50)ΔG=−485

As ΔG is - 485 (negative), the reaction will be spontaneous.

b) at 80°C

ΔG=ΔH−TΔSΔG=15−(80×50)ΔG=−3985

As ΔG is -3985 (negative), the reaction will be spontaneous.

Thus, the reaction will be spontaneous at 10°C and 80°C.

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