Determine at what temperature the reaction N2 3H2 -> 2NH3 will be product favored. delta H = -93 kJ/mol, delta S = -198 J/K mol?

The answer that I got is 469.7 K < T. Is this correct? Thank you for any help!

2 Answers
Apr 29, 2018

To determine thermodynamic favorability, we assume a reaction is thermodynamically favorable when #DeltaG<0#.

Recall,

#DeltaG = DeltaH - TDeltaS#

Given,

#DeltaH = (-93"kJ")/"mol"#

#DeltaS = (-198"J")/"K"#

#DeltaG = 0# (this is the threshold a process becomes thermodynamically favorable)

Hence,

#=> T = (DeltaH)/(DeltaS) approx 470"K"#

is the temperature at which the unfavorable entropy of this process is overcome.

The more we lower this temperature, the more the reaction will favor products.

Apr 29, 2018

This process is exothermic, so higher temperatures make it harder to work with in the industry.

We want the temperature for when the reaction will CEASE to be product-favored, and not when it will become spontaneous. That is not trivial, if I am to take your question as-written.

I get that #bb(T < "469.7 K")# (and not #T > "469.7 K"#). Above this temperature, the reaction STOPS being product favored, so we want the temperature to be lower to favor the products, not higher.


#K# varies with temperature, so we cannot simply say that #cancel(DeltaG)^(0) = DeltaG^@ + RTlncancel(Q)^(K)# and solve for #T#, because

#T = -(DeltaG^@)/(RlnK)#

which goes to infinity if #K = 1# as the question would be implying. Here are two ways to proceed.

METHOD 1

If #K = 1#, #DeltaG^@ = 0#, so

#0 = DeltaH^@ - TDeltaS^@#

As a result,

#T = (DeltaH^@)/(DeltaS^@) = (-"93 kJ/mol")/(-"0.198 kJ/mol"cdot"K"#

#=# #"469.7 K"#

We would have to recognize that since #DeltaH^@ < 0#, the higher temperature would disfavor the products. Therefore, #color(blue)(T_(max) = "469.7 K")# (not #T_(min)#).

METHOD 2

In this longer method, we could find #DeltaG^@#, and then use it to find #K_1#. Then, we could find #T# for which #K_2 = 1#. If this way works, we should get the same answer.

I would first find #DeltaG^@# at #"298.15 K"# (which is nonzero), and then find #K_1# at #"298.15 K"#:

#DeltaG^@ = DeltaH^@ - TDeltaS^@#

#= -"93 kJ"/"mol" - "298.15 K" cdot (-"0.198 kJ/mol"cdot"K")#

#= -"33.97 kJ/mol"#

The equilibrium constant at this temperature would then be:

#K_1 = e^(-(DeltaG^@)/(RT))#

#= "exp"[-(-"33.97 kJ/mol")/("0.008314472 kJ/mol"cdot"K" cdot "298.15 K")]#

#= e^(13.7)#

...Looks like it's VERY product-favored at #"298.15 K"# already... we can only go down in thermodynamic favorability for the forward reaction at higher temperatures for this exothermic process.

Then, I would use the van't Hoff equation to find what #T_2# temperature would lead to #K_2 = 1#, if we wanted to find the MAXIMUM temperature that it stays product-favored.

Again, since the reaction is exothermic, higher temperatures would STOP favoring the products and lower #K#, so we expect #T_2 > T_1#.

Assuming #DeltaH^@# stays constant in the temperature range:

#ln (K_2/K_1) = -(DeltaH^@)/R[1/T_2 - 1/T_1]#

#(Rln(1//K_1))/(-DeltaH^@) = 1/T_2 - 1/T_1#

#color(blue)(T_2) -= T_(max) = 1/((Rln(1//K_1))/(-DeltaH^@) + 1/T_1)#

#= 1/(("0.008314472 kJ/mol"cdot"K"cdotln(e^(-13.7)))/(-(-"93 kJ/mol")) + 1/"298.15 K")#

#=# #color(blue)("469.7 K")#

Above this temperature, the reactants become favored, so we want #T < "469.7 K"#.