Using data in table below and S °(C2H6(g))= 229.5 J·mol-1·K-1, calculate ÄfS° for C2H6(g) in J·mol-1·K-1.

Compound S °, J·mol-1·K-1 Compound S °, J·mol-1·K-1
C(s) 5.69 H2(g) 130.6

I am sorry , last time i did not show my working. thanks for helping me .

Delta S= Delta S product - Delta S reactant
and i did according to the formula that i got S =565.48
Is that correct

To calculate ΔfS° for C2H6(g), you need to use the formula:

ΔfS° = Σ(nΔS° products) - Σ(nΔS° reactants)

First, you need to determine the coefficients (n) for the compounds in the reaction:

C2H6(g) → C(s) + 3H2(g)

Since C2H6(g) is the product, the coefficient is 1. The reactants, C(s) and H2(g), have coefficients of 1 and 3, respectively.

Now, you can calculate ΔfS°:

ΔfS° = (1 * S°(C(s))) + (3 * S°(H2(g))) - (1 * S°(C2H6(g)))

Given:

S°(C(s)) = 5.69 J·mol-1·K-1
S°(H2(g)) = 130.6 J·mol-1·K-1
S°(C2H6(g)) = 229.5 J·mol-1·K-1

Plugging in the values:

ΔfS° = (1 * 5.69 J·mol-1·K-1) + (3 * 130.6 J·mol-1·K-1) - (1 * 229.5 J·mol-1·K-1)

Calculating:

ΔfS° = 5.69 J·mol-1·K-1 + 391.8 J·mol-1·K-1 - 229.5 J·mol-1·K-1

ΔfS° = 167.99 J·mol-1·K-1

Therefore, the value for ΔfS° for C2H6(g) is 167.99 J·mol-1·K-1.

To calculate the change in standard entropy (ΔfS°) for C2H6(g), we need to use the formula:

ΔfS° = ΣnS°(products) - ΣmS°(reactants)

First, let's determine the number of moles (n) for the product C2H6(g). Since there is only one mole of C2H6(g) in the equation, n = 1.

Next, we determine the number of moles (m) for the reactant H2(g). Again, there is only one mole of H2(g) in the equation, so m = 1.

Now, let's substitute the values into the formula:

ΔfS° = (n × S°(C2H6(g))) - (m × S°(H2(g)))
= (1 mol × 229.5 J·mol-1·K-1) - (1 mol × 130.6 J·mol-1·K-1)
= 229.5 J·mol-1·K-1 - 130.6 J·mol-1·K-1
= 98.9 J·mol-1·K-1

Therefore, ΔfS° for C2H6(g) is 98.9 J·mol-1·K-1.