How to Calculate Entropy Changes at Non-Standard Conditions

Question

I am trying to calculate the values of $\Delta H$ and $T\Delta S$ for the reaction taking place at $\pu{800 ^\circ C}$

$$ \ce{CO2 + H2O + 2CH4 -> 3CO + 5H2} $$

I calculated $\Delta H$ to be $\pu{\approx 480 kJ mol^{-1}}$ using Hess’ Law, i.e., by taking the reactants down to $\pu{ 25 ^\circ C}$ and calculating the enthalpy change, performing the reaction at standard conditions, and then taking the products back up to $\pu{800 ^\circ C}$.

However, I am having trouble with $T\Delta S$. Can I take the same approach as I did for the enthalpy change, taking the reactants/products between $\pu{25 ^\circ C}$-$\pu{800 ^\circ C}$ and calculating the entropy change at $\pu{25 ^\circ C}$? Then multiplying by the temperature in Kelvin?

I tried using the Shomate equation expression below for the reactants/products (taken from NIST), where A-G are Shomate equation coefficients; however, I’m not getting the correct answer of $\pu{\approx -80 kJ mol^{-1}}$, so unsure if this is the correct approach:

$$ \begin{align} \Delta S = &A \ln{(T_2-T_1)} + B(T_2-T_1) + \dfrac{C}{2}(T_2^2-T_1^2)\\ &+ \dfrac{D}{3}(T_2^3-T_1^3) - \dfrac{E}{2(T_2^2-T_1^2)} + G \end{align} $$

Answer 1

You maybe applying the Shomate equation incorrectly:

$$ \ln{T_2} - \ln{T_1} = \ln{\dfrac{T_2}{T_1}} $$

$$ \dfrac{1}{T_2^2} - \dfrac{1}{T_1^2} = \dfrac{T_1^2-T_2^2}{T_1^2T_2^2} $$

The correct application of the Shomate equation would be:

$$ \begin{align} S ^\circ &= A \ln{(T)} + B(T) + \dfrac{C}{2}(T^2)\\ &+ \dfrac{D}{3}(T^3) - \dfrac{E}{2(T^2)} + G\\ \implies \Delta S ^\circ &= A \ln{\dfrac{T_2}{T_1}} + B(T_2-T_1) + \dfrac{C}{2}(T_2^2-T_1^2)\\ &+ \dfrac{D}{3}(T_2^3-T_1^3) - \dfrac{E}{2}\left( \dfrac{T_1^2-T_2^2}{T_1^2T_2^2}\right) + G \end{align} $$

How to Calculate Entropy Changes at Non-Standard Conditions

Do note that conditions are still standard, and not non-standard, as indicated by the $\circ$ superscript in the above equations.