We rearrange the equation to give us the following: and finally, we factor out the heat quantity (Q) to yield the final form of the equation: What is the the following? ΔS given = -37.2 kJ/mol ΔSfinal = -32.2 kJ/mol ΔSinitial = SFinal SInitital ΔS – -37.2 kJ/mol - -32.2 kJ/mol = -5 kJ/mol What is the entropy change, given the following information? Q = 1.1 104 J/mol x T1 = 12° C T2 = 5° C T1 = 12° C = 285 K T2 = 5° C = 278 K = 1.1 x 104 J/mol (3.60 x 10-3 K-1 3.51 x 10-3K-1) ΔS – = 1.1 104 J/mol (0.09 10-3 K-1) x x ΔS = 0.099 x 10-1 K-1 ΔS = 1.0 J/Kmol ΔS
We rearrange the equation to give us the following: and finally, we factor out the heat quantity (Q) to yield the final form of the equation: What is the ΔS given the following? ΔSfinal = -37.2 kJ/mol ΔSinitial = -32.2 kJ/mol ΔS = SFinal – SInitital -37.2 kJ/mol - -32.2 kJ/mol = -5 kJ/mol What is the entropy change, given the following information? Q = 1.1 x 104 J/mol T1 = 12° C T2 = 5° C T1 = 12° C = 285 K T2 = 5° C = 278 K ΔS = 1.1 x 104 J/mol (3.60 x 10-3 K-1 – 3.51 x 10-3K-1) ΔS = 1.1 x 104 J/mol (0.09 x 10-3 K-1) ΔS = 0.099 x 10-1 K-1 ΔS = 1.0 J/Kmol
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