For the equilibrium A + B ⇌ C + D, K was measured to be 1.51 × 10⁻² at 25°C and 135.0 at 600°C. What is ∆S° for the reaction? Assume ∆H° and ∆S° are temperature independent.

Answer 1

Great question! This one requires a couple of different calculations, and two main equations -- the Van't Hoff Equation and the Gibbs Free Energy equation.

Let's start with the first part -- we need to calculate delta H. We can use the Van't Hoff equation to do that. (Don't forget to convert to Kelvin!)

ln(K1/K2) = -deltaH / R * (1/T2 - 1/ T1)

With your values...

ln (1.51*10^-2 / 135) = - deltaH / 8.314 J/mol*K * ( 1/873 - 1/298)

which...after all the math, works out to a delta H of -34224 J/mol, or -34.2 kJ/mol.

Now, since we have that, we can use the Gibbs Free Energy equation to solve for our delta S value! The only problem is, without having any idea what delta G for the reaction is, it's a little bit tough to do that. But, we can work around that by using the value of delta G of zero, because we know that when K=1, delta G is zero! The best way to figure out what the temperature is when delta G is zero is to first solve for it using the Van't Hoff equation again!

So again...

ln(K1/K2) = -deltaH / R * (1/T2 - 1/ T1)

Let's just use the T2 and K2/ from above, then allow K1 to be 1 (perfect equilibrium) and solve for T1!

ln(1/135) = -34224 / 8.314 * ( 1/873 - 1/T1)

which...after all the new math, gives a temperature of T1 as 427.9K.

Finally, we get to figure out delta S, using the Gibbs Free Energy equation, delta G = delta H - T * delta S.

Since we know that we're at equilibrium, delta G = 0, delta H we calculated above, and we have the T from the most recent equation. So putting it together....

0 = -34224 J/mol - 427.9K * delta S

Which gives us a value of 80.15 J/mol*K for delta S!

Hope that helps :)

-Niral