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When gaseous nitrogen and hydrogen are converted to gaseous ammonia, [delta]G0=

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Endeavor:
'When gaseous nitrogen and hydrogen are converted to gaseous ammonia, [delta]G0= 16.64 kJ/mol.
The chemical equation is
N2(g) + 3H2(g) ----> 2NH3(g)

Using the S0 values in Table 19.2, calculate the [delta]H0f for the formation of ammonia."

The S0(J/(K*mol)) for N2(g) = 191.5
for H2(g) = 130.6
for NH3(g) = 192.5

So I convert these values to kJ/(K*mol) and multiply them by their coefficients:
for N2(g) = 0.1915 kJ/(K*mol)
for 3H2(g) = 0.3918 kJ/(K*mol)
for 2NH3(g) = 0.385 kJ/(K*mol)

[delta]S0 = -0.1983 kJ(K*mol)

[delta]G = [delta]H0 - T * [delta]S0
if T = 273K, then
[delta]H0 = -37.4959

....but now I'm getting lost in all these numbers. HELP

charco:
You seem to be going about it the correct way i.e. calculating the deltaS value for the equation and substituting it into the Gibbs free energy equation...

However, don't forget that standard state is NOT STP i.e. the T value should be 298K (unless quoted otherwise), not 273K

OK I was being lazy....

calculate absolute entropy for the LHS  and absolute entropy for the RHS and entropy change will equal RHS - LHS (total final entropy minus original entropy)

In this case [2 x 192.5] - [191.5 + (3 x 130.6)]

= 385 - 583.3 =  -198.3 ( correctly stated in kJmol-1 as -0.1983)

delta G = delta H - T delta S

delta H = delta G + T delta S

delta H = 16.64 + (298 x -0.1983)

delta H = 16.64 - 59.09 = - 42.45 kJmol-1

Endeavor:
Okay.  But how do I find the [delta]H0f for the formation of ammonia??

charco:
if everything is calculated at standrd conditions then this IS delta Hfº

Endeavor:
that can't be...because the answer in my book is -75.2 kJ/mol

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