In the previous question, you had to calculate the the standard Free Energy Chan

Question

In the previous question, you had to calculate the the standard Free Energy Change (?Go) in order to solve for the equilibrium constant, K, for the reaction:

N2(g) + 3H2(g) ? 2NH3(g)

This is the Free Energy measured under standard conditions, when the reaction is started with 1.0 M of each of the three gases present. Calculate the non-standard Free Energy change (?G) at 298 K, given the following non-standard initial partial pressures of the three gases. Answer in kJ.
?Gfo = -16.6 kJ/mol for NH3 (g) at 298 K

? G =

Explanation / Answer

?G=?Gfo+RT lnQ

Let's find the ?Gfo value to use by multiplying it by 2 because there are 2 moles of NH3(g). Then, convert it to J just because it's easier to calculate.
?Gfo= -16.6 kJ/mol *2= -33.2 kJ/mol= -33,200 J/mol

And we know that
R=8.314 J/mol*K
T=298 K

Now all we need to do is to solve for Q (reaction quotient).
We can do so by using the partial pressures of reactants and product. You can just look up the formula but here are the calculations:
Q=(1.9)^2/[1.0*0.04^3]= 56406.25

Now that we have the values for ?Gfo, R, T, and Q, we will be able to plug them in the original equation (this one: ?G=?Gfo+RT lnQ) and solve for ?G. Finally, remember to convert everything back to kJ, since that's what the question asks for.

?G= -33,200+8.134*298*ln 56406.25 = -6094.53 J = -6.09453 kJ

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