Question

Given the following equations, calculate the ΔH of reaction. SHOW ALL YOUR WORK AND CALCULATIONS FOR FULL CREDIT

Answer 1

The ΔH of the reaction is -769 kJ/mol.

To calculate the ΔH of the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l), you can use Hess's law by manipulating and combining the given equations:

4NH3​(g)+5O2​(g)→4NO(g)+6H2​O(l)

2NO(g)→N2​(g)+O2​(g)

2NO(g)+2NH3​(g)→2N2​(g)+2H2​O(l)

Now, sum these equations to get the overall reaction:

4NH3​ (g)+5O2(g)+2NO(g)+2NH3​(g)→4NO(g)+6H2​O(l)+N2​(g)+O2(g)+2N2​(g)+2H2​O(l)

Simplify the equation:

4NH3​(g)+5O2 (g)→4NO(g)2O(l)+N2(g)+O2​(g)

Now, sum the enthalpy changes:

ΔHtotal​=ΔH1​+ΔH2​+ΔH3

ΔHtotal​=−183.5kJ/mol+(−91.9kJ/mol)+(−493.6kJ/mol)

ΔHtotal =−769kJ/mol

Therefore, the ΔH of the reaction is -769 kJ/mol.