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What is the value of the equilibrium constant at 655 K for each of the following reactions?

#2NO_2(g)harrN_2O_4(g)#
#Br_2(g)+Cl_2(g)harr2BrCl(g)#

I figured out how to calculate #DeltaG^0# using #DeltaH^0# and #DeltaS^0#, but I am unable to determine the value for #DeltaG# because it is not at STP. Do I just plug in #DeltaG^0# to the #-RTlnK# equation to find the value of K, or do I have to adjust something for its not being at STP?

Answer 1

Henry Antrim

#K approx 1.67#

Yes! You're correct, I'll do the first one.

#2NO_2(g) rightleftharpoons N_2O_4(g)#

#DeltaG^0 approx -2.8kJ#

I did that per the usual method using a table in my text.

#-2.8*10^3 J = -(8.314J)/(mol*K)* (655K) * lnK# #therefore K approx 1.67#

This is reasonable because, I did a quick check and noticed this reaction has favorable enthalpy but unfavorable entropy. Thus, it will be nonspontaneous at high temperatures. Thus, not much product will form relative to standard conditions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.