ΔHo for a reaction F2 + 2HCl → 2HF + Cl2 is given as -352.8 KJ. ∆Hf0 for HF is –268.3KJ mol-1, then ∆Hf0 of HCl would be

F2 + 2HCl → 2HF + Cl2 , ΔHo =-352.8KJ

ΔHo for a reaction F2 + 2HCl → 2HF + Cl2 is given as -352.8 KJ. ∆Hf0 for HF is –268.3KJ mol-1, then ∆Hf0 of HCl would be

F2 + 2HCl → 2HF + Cl2 , ΔHo =-352.8KJ

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ΔH^o for a reaction F₂ + 2HCl → 2HF + Cl₂ is given as -352.8 KJ. $∆ H_{f}^{0}$ for HF is –268.3KJ mol^-1, then $∆ H_{f}^{0}$ of HCl would be

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– 91.9kJmol^–1

– 183.8 kJmol^–1

– 22 kJmol^–1

880 kJmol^–1

answer is C.

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Detailed Solution

F₂ + 2HCl → 2HF + Cl₂ , ΔH^o =-352.8KJ

$\begin{array}{l} \Delta {H^o} = [2{\Delta _f}H(HF) + {\Delta _f}H(C{l_2})] - [{\Delta _f}H({F_2}) + 2{\Delta _f}H(HCl)]\\ - 352.8 = [2( - 268.3) + 0] - [0 + 2{\Delta _f}H(HCl)]\\ - 352.8 = - 536.6 - 2{\Delta _f}H(HCl) \end{array}$

${\Delta _f}H(HCl) = \frac{{ - 183.8}}{2} = - 91.9KJ/mole$

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