Question

Given the values of ?H?rxn, ?S?rxn, and T below, determine ?Suniv.

a)?H?rxn=? 133kJ , ?S?rxn= 263J/K , T= 291K .

b)?H?rxn= 133kJ , ?S?rxn=? 263J/K , T= 291K .

c)?H?rxn=? 133kJ , ?S?rxn=? 263J/K , T= 291K .

d)?H?rxn=? 133kJ , ?S?rxn=? 263J/K , T= 553K .

Answer 1

NOTE: In general, the change in temperature of surroundings could be considered 0. There will be no change since the surroundings are huge ( for example, the ocean, the atmosphere, a mountain, and so on).

Assume T = constant

If T of surrounding is constant

Let Qsurrounding = -Hrxn

The heat gained/lost by the surrounding is equal to the heat lost/gained by the system

Entropy is given by

S = Q/T = -H/T

Let us calculate the entropy of the surroundings for each case

a)

S surrounding = -H/T = -(-133 kJ) / 291 K = 0.457 kJ/K

S universe = S system + S surrounding = 0.263 kJ/K + 0.457 kJ/K = 0.72 kJ/K or 720 J/K

b)

S surrounding = -H/T = -(133 kJ) / 291 K = -0.457 kJ/K

S universe = S system + S surrounding = -0.263 kJ/K -0.457 kJ/K = -0.72 kJ/K or -720 J/K

c)

S surrounding = -H/T = -(-133 kJ) / 291 K = 0.457 kJ/K

S universe = S system + S surrounding = -0.263 kJ/K + 0.457 kJ/K = 0.194 kJ/K or 194 J/K

d)

S surrounding = -H/T = -(-133 kJ) / 553 K = +0.241 kJ/K

S universe = S system + S surrounding = -0.263 kJ/K + 0.241 kJ/K = 0.022 kJ/K or 22 J/K

Answer 2

a). Srxn=263J/K = 0.263kJ/K
?H