Calculate the entropy of formation (J/K mol) of CS2(l). The value of T delta S^0 for the equation as written is -16.7774 KJ/mol. (the given standard entropies are: O2= 205.138, CO2= 213.7, SO2= 248.22)
CS2(l) + 3 O2(g) --> CO2(g) + 2 O2(g)
At what temperature is the value of
T delta S^0 equal to -16.7774 KJ/mol.?
I believe you mean SO2 as the second product on the right.
Once you know T, you can set the difference of the standard entropies of the reactants and products equal to the delta S of the reaction, and solve for the entropy of formation of CS2(l)
To calculate the entropy of formation (ΔS°f) of CS2(l), you need to use the given TΔS° value and the standard entropies of the reactants and products involved in the chemical equation.
The equation is:
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)
Step 1: Calculate the change in entropy (ΔS) for the equation.
Since the equation is already balanced, you can use the stoichiometric coefficients as the coefficients for ΔS calculations.
ΔS = Σn(prod)S°(prod) - Σn(react)S°(react)
ΔS = [1 × S°(CO2)] + [2 × S°(SO2)] - [1 × S°(CS2)] - [3 × S°(O2)]
Step 2: Substitute the given standard entropies into the equation.
ΔS = [1 × 213.7 J/(K mol)] + [2 × 248.22 J/(K mol)] - [1 × ?] - [3 × 205.138 J/(K mol)]
Step 3: Calculate the unknown standard entropy of CS2(l).
To calculate the unknown standard entropy of CS2(l), rearrange the equation:
ΔS = -16.7774 kJ/mol = [1 × 213.7 J/(K mol)] + [2 × 248.22 J/(K mol)] - [1 × ?] - [3 × 205.138 J/(K mol)]
Remember to convert the given TΔS° value from kJ/mol to J/(K mol).
Step 4: Solve for the unknown standard entropy of CS2(l).
-16.7774 × 10^(3) J/(K mol) = 213.7 J/(K mol) + 496.44 J/(K mol) - CS2(l) - 615.414 J/(K mol)
CS2(l) = 213.7 J/(K mol) + 496.44 J/(K mol) - 615.414 J/(K mol) - (-16.7774 × 10^(3)) J/(K mol)
CS2(l) = 213.7 J/(K mol) + 496.44 J/(K mol) - 615.414 J/(K mol) + 16.7774 × 10^(3) J/(K mol)
Therefore, the entropy of formation (ΔS°f) of CS2(l) is the calculated value of CS2(l) from the above expression, which is in J/(K mol).