# 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?

## "Predict whether or not this reaction will be spontaneous at this temperature" $4 N {H}_{3}$ (g) + $3 {O}_{2}$ (g) $\rightarrow 2 {N}_{2}$ (g) + $6 {H}_{2} O$(g) $\setminus \Delta H = - 1267$ kJ A) $\setminus \Delta {S}_{s u r r}$ =+12.67 kJ/K, rxn not spontaneous B) $\setminus \Delta {S}_{s u r r}$ = -12.67 kJ/K, rxn spontaneous C) $\setminus \Delta {S}_{s u r r}$ = +50.4 kJ/K, rxn not spontaneous D) $\setminus \Delta {S}_{s u r r}$ = +3.18 KJ/K, rxn spontanous E) $\setminus \Delta {S}_{s u r r}$ -3.18 kJ/K, it is not possible to predict the spontaneity of this rxn without more information

Apr 28, 2017

C)

#### Explanation:

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

$\setminus \Delta S = - \frac{\setminus \Delta H}{T} = - \frac{- 1267 k J}{398 K} = + 3.18 \frac{k J}{K}$

$\setminus \Delta G = - T \setminus \Delta S = - \left(398 K\right) \setminus \cdot \left(+ 3.18 \frac{k J}{K}\right) \setminus \approx - 1.27 \setminus \cdot {10}^{3} k J$
as $\setminus \Delta G \setminus < 0$; rxn spontaneous

Apr 28, 2017

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

#### Explanation:

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.