Question

Complete the following:

a. Using standard heats of formation, determine ΔH for the following reaction:

4NH₃(g) + 5O₂(g) →4NO(g) + 6H₂O(l)

b. Estimate the difference between ΔH_reaction and ΔE_reaction (in kJ/mol) for the reaction:

C₄₀H₈₀(s) + 60O₂(g) →40CO₂(g) + 40H₂O(s) at 298 K

Answer 1

Answer -

a)

Standard values,

ΔHf (NH3 (g)) = - 46.11 kJ/mol

ΔHf (O2 (g)) = 0 kJ/mol

ΔHf (NO(g)) = 90.25 kJ/mol

ΔHf (H2O(l)) = - 285.83 kJ/mol

ΔHrxn = ?

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) [BALANCED]

We know that,

ΔHrxn = np * ΔHf (products) - nr * ΔHf (reactants)

ΔHrxn = 4 * ΔHf (NO(g)) + 6 * ΔHf (H2O(l)) - 4 * ΔHf (NH3(g)) - 5 * ΔHf (O2(g))

ΔHrxn = (4*90.25 kJ/mol) + (6*-285.83 kJ/mol) - (4*- 46.11 kJ/mol)-(5*0 kJ/mol)

ΔHrxn = (4*90.25 kJ/mol) - (6*285.83 kJ/mol) +(4*46.11 kJ/mol)

ΔHrxn = -1169.54 kJ/mol [ANSWER]

b)

Given,

Temperature = 298 K

C40H80(s) + 60 O2(g) → 40 CO2(g) + 40 H2O(s) [BALANCED]

ΔH - ΔE = ?

We know that,

H = E + PV

Also,

PV = nRT

So,

ΔH = ΔE + ΔnRT

ΔH - ΔE = ΔnRT

where,

Δn = (moles of products - moles of reactants)

R = Gas Constant (8.314 J/Kmol)

T = Temperature

Put the values,

ΔH - ΔE = (40 + 40 -1 -60) * 8.314 J/K.mol * 298 K

ΔH - ΔE = 47073.868 J/mol

Now,

1 J/mol = 0.001 kJ/mol

So, 47073.868 J/mol = 47073.868 * 0.001 kJ/mol

47073.868 J/mol = 47.07 kJ/mol

ΔH - ΔE = 47.07 kJ/mol [ANSWER]