The standard enthalpy of formation (Δf H₃₀₀⁰) at 1 bar and 300 K for the formation of CF₂ ClCF₂ Cl(g) from… · ABC26PH0842

Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Thermodynamics – General · Exam: JAM 2001-2022 · Marks: 2 · Difficulty: Medium

The standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}_{300}^{0}\right)$ at 1 bar and 300 K for the formation of $\mathrm{CF}_{2} \mathrm{ClCF}_{2} \mathrm{Cl}(\mathrm{g})$ from its constituent elements in the standard state is $-900 \mathrm{kJ} \mathrm{mol}^{-1}$. Given $\mathrm{R}=8.3 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}$, the standard internal energy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}_{300}^{0}\right)$ at the same pressure and temperature is
(a)$-905 \mathrm{kJ} \mathrm{mol}^{-1}$
(b)$-895 \mathrm{kJ} \mathrm{mol}^{-1}$
(c)$895 \mathrm{kJ} \mathrm{mol}^{-1}$
(d)$905 \mathrm{kJ} \mathrm{mol}^{-1}$
Answer
Answer: B ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
2C(s) + 2F₂(g) + Cl₂(g) → CF₂ClCF₂Cl(g): Δn_gas = 1 − 3 = −2. ΔU = ΔH − Δn RT = −900 + 2 × 8.3 × 300/1000 = −895 kJ mol⁻¹.

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