Question

For a particular reaction at 123.3 °C, Δ?=−443.64 kJ , and Δ?=874.77 J/K . Calculate ΔG for this reaction at 3.2 °C. kJ

Answer 1

Let's first see what is given in the problem

01.G at temp123.3 °C = -443.64KJ

02.s value at 123.3° C = 874.77j/mol = 0.87477 KJ

03.temp = 123.3 °C = (123.3 +273.15) = 396.45 K

to find

G at 3.2 °C ( 276.35 K )

We know the formula

G = H - TS

putting the values given in the problem

-443.64 kj = H - (396.45× 0.87477)

H = -443.64 + 346.80

H = -96.83 KJ

Using this value to calculate G at 3.2 °C (276.35 K)

G = H -TS

G = -96.83 - (276.35 × 0.87477)

G = -96.83 - 241.74

G= -338.57 KJ

So the answer is -338.57 Kj at given temp .

I hope this helps if you have any query or want more detailed explanation feel free to ask in the comment section below.

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