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

Mateo Aguilar · Accepted Answer

C)

Using the notes from class, this is what I came up with. #\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

Jeremiah Adams · Answer

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.

Cameron Andrews · Answer

undefined To determine the change in entropy of the surroundings ($ΔS_{surr}$), we need to consider the change in enthalpy ($ΔH$) and temperature ($T$) of the reaction. If $ΔH$ is negative (exothermic reaction) and $T$ is positive, $ΔS_{surr}$ will be negative.

Without the specific reaction and $ΔH$ value, we cannot calculate $ΔS_{surr}$ or determine spontaneity.