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### Topic: When gaseous nitrogen and hydrogen are converted to gaseous ammonia, [delta]G0=  (Read 9828 times)

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#### Endeavor

• Guest ##### When gaseous nitrogen and hydrogen are converted to gaseous ammonia, [delta]G0=
« on: April 15, 2005, 01:41:10 PM »
'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
« Last Edit: April 19, 2005, 04:58:15 AM by Mitch »

#### charco

• Guest ##### Re:I don't know how to solve this...
« Reply #1 on: April 15, 2005, 08:05:54 PM »
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
« Last Edit: April 15, 2005, 08:16:50 PM by charco »

#### Endeavor

• Guest ##### Re:I don't know how to solve this...
« Reply #2 on: April 17, 2005, 05:51:08 PM »
Okay.  But how do I find the [delta]H0f for the formation of ammonia??

#### charco

• Guest ##### Re:I don't know how to solve this...
« Reply #3 on: April 17, 2005, 05:59:27 PM »
if everything is calculated at standrd conditions then this IS delta Hfº

#### Endeavor

• Guest ##### Re:I don't know how to solve this...
« Reply #4 on: April 18, 2005, 11:29:23 AM »
that can't be...because the answer in my book is -75.2 kJ/mol

#### charco

• Guest ##### Re:I don't know how to solve this...
« Reply #5 on: April 19, 2005, 03:01:04 AM »
This is a strange answer as the quoted literature value is -46kJ mol-1

Seems like there are a couple of problems...

1. The value of deltaG given is incorrect - it should be negative!

now for the reaction that forma two moles of ammonia delta G = 2 * -16.64 = -33.28 kJ

T delta S = 298 * -0.1982 = -59.0636
deltaG = deltaH - T delta S
delta H = delta G + T delta S
delta H = -33.28 - 59.0636 = -92.34Kj FOR THE REACTION

Hence delta H formation = 92.34/2 = -46.2 kJmol-1

check out this full treatment of the Haber process

http://blue.butler.edu/~jkirsch/thermo.pdf

« Last Edit: April 19, 2005, 04:10:42 AM by charco »