Question

Calculate the net change in energy for the following reaction:
2Na(s) + 2HCl(g) → 2NaCl(s) + H2(g)
Given the following information:
Energy of sublimation of Na(s) = 97 kJ/mol
Bond energy of HCl = 427 kJ/mol
Ionization energy of Na(g) = 496 kJ/mol
Electron affinity of Cl(g) = –349 kJ/mol
Lattice energy of NaCl(s) = –778 kJ/mol
Bond energy of H2 = 432 kJ/mol

Answer 1

Answer:

-1078 kJ/mol.

Explanation:

Consider this reaction in three steps:

1. Break everything down into atoms: 2 Na(g) + 2 H(g) + 2 Cl(g);
2. Turn 2 Na(g) and 2 Cl(g) into 2 Na⁺(g) and 2 Cl⁻(g);
3. Combine everything to form 2 NaCl(s) and H₂(g).

Sources of enthalpy changes in the first step:

• Sublimate 2 Na(s);
• Break two H-Cl bonds.

That corresponds to an enthalpy change of

$\rm 2\times 97 + 2\times 427 = 1,048\;kJ\cdot mol^{-1}$.

Sources of enthalpy changes in the second step:

• Remove one electron for each of the two Na(g);
• Add one electron to each of the two Cl(g).

That corresponds to an enthalpy change of

$\rm 2\times 496 + 2\times (-349) = 294\; kJ\cdot mol^{-1}$.

Sources of enthalpy changes in the third step:

• Bring 2 Na⁺(g) and 2 Cl⁻(g) together to form 2 NaCl(s) (which corresponds to twice the lattice enthalpy of NaCl);
• Bring 2 H(g) together to form one H₂(g).

That corresponds to an enthalpy change of

$\rm 2\times (-778) + 2\times (-432) = -2,420\;kJ\cdot mol^{-1}$.

Take the sum of the enthalpy changes of the three steps to find the enthalpy change of the overall reaction:

$\rm 1,048 + 294 + (-2,420) = -1078\; kJ\cdot mol^{-1}$.