Using the following thermochemical data:
2Cr(s) + 6HF(g) $\to$ 0 2CrF₃(s) + 3H₂(g) $\quad$ $\Delta$ H° = -691.4 kJ/mol
2Cr(s) + 6HCl(g) $\to$ 2CrCl₃(s) + 3H₂(g) $\quad$ $\Delta$ H° = -559.2 kJ/mol
Calculate $\Delta$ H° for the following reaction:
CrF₃(s) + 3HCl(g) $\to$ CrCl₃(s) + 3HF(g)

Question

Using the following thermochemical data:
2Cr(s) + 6HF(g)

\to

0 2CrF₃(s) + 3H₂(g)

\quad

\Delta

H° = -691.4 kJ/mol
2Cr(s) + 6HCl(g)

\to

2CrCl₃(s) + 3H₂(g)

\quad

\Delta

H° = -559.2 kJ/mol
Calculate

\Delta

H° for the following reaction:
CrF₃(s) + 3HCl(g)

\to

CrCl₃(s) + 3HF(g)

Quizplus · Accepted Answer

1. Write out the given equation:
Cr(s) + 6HCl(g) → CrCl3(s) + 3H2(g)
2. Flip the first equation and change the sign of ΔH for the first equation so that it can cancel with the products in the above equation:
2CrCl3(s) + 3H2(g) → 2Cr(s) + 6HCl(g)         ΔH° = 559.2 kJ/mol
2Cr(s) + 6HF(g) → 2CrF3(s) + 3H2(g)         ΔH° = -691.4 kJ/mol
3. Combine the above two equations so that the reactants will give the products of the given reaction:
2CrCl3(s) + 6HF(g) → 2CrF3(s) + 6HCl(g)         ΔH° = ?
4. Add the enthalpies of the two equations to find the enthalpy for the third equation:
(2Cr(s) + 6HF(g) → 2CrF3(s) + 3H2(g)) × 2
ΔH° = 2(-691.4 kJ/mol) = -1382.8 kJ/mol
(2CrCl3(s) + 3H2(g) → 2Cr(s) + 6HCl(g)) × 2
ΔH° = 2(559.2 kJ/mol) = 1118.4 kJ/mol
2CrCl3(s) + 6HF(g) → 2CrF3(s) + 6HCl(g)
ΔH°= -1382.8 kJ/mol + 1118.4 kJ/mol = -264.4 kJ/mol

Therefore, the answer is C) -264.4 kJ/mol.

Using the Following Thermochemical Data: 2Cr(s)+ 6HF(g) →\to→ 0 2CrF3(s)+ 3H2(g)

Using the Following Thermochemical Data:
2Cr(s)+ 6HF(g) $\to$ 0 2CrF₃(s)+ 3H₂(g)