Answer to Question #96821 in General Chemistry for Lara Casteel

Question

Given the following enthalpy change values for the reactions below:

A + 2B → C ; ΔrH = -228.3 kJ mol-1

E + F → 2C + D ; ΔrH = 493.9 kJ mol-1

½ E → G + 2B ; ΔrH = 314.3 kJ mol-1

Calculate ΔrH for the following reaction (in kJ mol-1) using Hess's Law:

2A + D → F + 2G

Answer 1

First, we have to multiply first equation with 2:

2A + 4B = 2C (2*(-228.3kJ/mol) = -456.6kJ/mol);

Further, we have to subtract the second equation from the first multiplied with 2:

2A + 4B - E - F = 2C - 2C - D (-456.6kJ/mol - 493.9kJ/mol = -950.5kJ/mol);

Now, we have to multiply the third equation with 2:

E = 2G + 4B (2*314.3kJ/mol = 628.6kJ/mol);

And finally, we have to combine last two modified equations:

2A + 4B - E - F + E = 2G + 4B - D, or

2A + D = 2G + F (628.6kJ/mol - 950.5kJ/mol = -321.9kJ/mol);

ΔrH = 2*(-228.3) - 493.9 + 2*314.3 = -321.9 kJ/mol.