In a given process, the entropy change of the surroundings (DeltaS_surr) is 4.3

Question

In a given process, the entropy change of the surroundings (DeltaS_surr) is 4.3 J/K and the entropy change of the system (DeltaS) was found to be 25.7 J/K at a given temperature. Calculate the entropy change of the universe and determine whether the process is thermodynamically favorable or spontaneous. Calculate DeltaS_surr for the following reactions at 25degree C and I atm. a. C_3H_8(g) + 5O_2(g) rightarrow 3CO_2(g) + 4H_2O(1) DeltaHdegree = -2221 kj b. 2NO_2(g) rightarrow 2NO(g) + O_2(g) DeltaHdegree = 112 kj

Explanation / Answer

Entropy change of universe = entropy change of surroundings + entropy change of system= 4.3+25.7= 30 J/k. Since entroy change is +ve, the process is thermodynamically possibe.

1. for the reaction, C3H8+ 5O2------> 3CO2 +4H2O deltaG= -2284 Kj

deltaG= deltaH- TdeltaS

TdeltaS= deltaH- deltaG= -2221+2284= 63 Kj

deltaS= 63/298.15 Kj/K=0.211 Kj/K

b) for the reaction 2NO2(g)-----------2NO + O2 delta G= 70.48 Kj

deltaH= 112 Kj , TdeltaS= 112-70.48= 41.52 Kj

deltS= 41.52/298.15=0.139 Kj/K

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