Consider the following reaction; use the information here to determine the value of $\DeltaS_(surr)$ at 398 K ... Will this reaction be spontaneous at this temperature?

Question

"Predict whether or not this reaction will be spontaneous at this temperature"
$4NH_3$ (g) + $3O_2$ (g) $rarr2N_2$ (g) + $6H_2O$ (g)
$\DeltaH=-1267$ kJ

A) $\DeltaS_(surr)$ =+12.67 kJ/K, rxn not spontaneous
B) $\DeltaS_(surr)$ = -12.67 kJ/K, rxn spontaneous
C) $\DeltaS_(surr)$ = +50.4 kJ/K, rxn not spontaneous
D) $\DeltaS_(surr)$ = +3.18 KJ/K, rxn spontanous
E) $\DeltaS_(surr)$ -3.18 kJ/K, it is not possible to predict the spontaneity of this rxn without more information

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Answer 1 · 2017-04-28T00:31:28

C)

This is what I got based off of class notes...

$\DeltaS=-(\DeltaH)/T=-(-1267kJ)/(398K)=+3.18(kJ)/K$

$\DeltaG=-T\DeltaS=-(398K)\cdot(+3.18(kJ)/K)\approx-1.27\cdot10^3kJ$
as $\DeltaG\lt0$ ; rxn spontaneous

Answer 2 · 2017-04-28T01:28:00

The correct answer is D) $ΔS_text(surr) = "+3.18 kJ/K"$ ; rxn spontaneous.

$ΔS_text(surr) = "-"(ΔH_text(sys))/T = "-""-1267 kJ"/"398 K" = "+3.18 kJ/K"$

For a reaction to be spontaneous, we must have $ΔG <0$ .

$ΔG = ΔH -TΔS$

The sign of $ΔH$ is negative, and the sign of $ΔS$ is positive, because the number of moles is increasing.

Thus, the sign of $ΔG$ is negative at all temperatures, and the reaction is spontaneous at all temperatures.