Question

The standard heats of formation for CCl_4(g), H₂O_(g), CO_2(g), and HCl_(g) are −25.5, −57.8, −94.1 and −22.1 kcal mol⁻¹, respectively.

∆H for the reaction

\[\ce{CCl4_{(g)} + 2H2O_{(g)} -> CO2_{(g)} + 4HCl_{(g)}}\] at 298 K

Options

• −265.8 kcal

• −199.5 kcal

• −32.9 kcal

• −41.4 kcal

Answer 1

−41.4 kcal

Explanation:

At 298 K, ∆H is ∆H⁰, i.e. standard heat of formation

For, \[\ce{CCl4_{(g)} + 2H2O_{(g)} -> CO2_{(g)} + 4HCl_{(g)}}\]

∆H⁰ = ?

`∆"H"_"Reaction"^0 = ∆"H"_"Products"^0 - ∆"H"_"Reactants"^0`

= `[∆"H"_("f"_(CO_2))^0 + 4 xx ∆"H"_("f"_(HCl))^0] - [∆"H"_("f"_(CCCl_4))^0  + 2 xx ∆"H"_("f"_(H_2O"))^0]`

= [1 mol × (−94.1 kcal mol⁻¹) + 4 mol × (−22.1 kcal mol⁻¹)] − [1 mol × (−25.5 kcal mol⁻¹) + 2 mol × (−57.8 kcal mol⁻¹)]

= −41.4 kcal