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Thermodynamics : Heats of Formation

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For the reaction CH4 (g) + N2 (g) + 164 kJ 6 HCN(g) + NH3 (g) at 25degrees C and 1 atm of pressure
Ã„Go = 159 kJ. Calculate Ã„So at 25degrees C.

Ã„Hfo for CH4 (g) = -74.77 kJ/mol and Ã„Hfo for
NH3 (g) = -46.09 kJ/mol. What is Ã„Hfo for HCN(g) ?

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For the reaction CH4 (g) + N2 (g) + 164 kJ 6 HCN(g) + NH3 (g) at 25degrees C and 1 atm of pressure
Ã„Go = 159 kJ. Calculate Ã„So at 25degrees C.

Looking at your reaction, you know the reaction required 164 kJ of heat. This is your Î”H value. You know Î”G = Î”H - TÎ”S, so you can plug in the given values to solve:

Ã„Hfo for CH4 (g) = -74.77 kJ/mol and Ã„Hfo for
NH3 (g) = -46.09 kJ/mol. What is Ã„Hfo for HCN(g) ?

To find the heat of formation, you need to find the products minus the reactants. Remember the heat of formation of N2 is zero because it is in its natural state:

The heat of formation of HCN is 130 kJ/mol.

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