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F(2)(g) +2HCI(g) rarr 2HF(g) +CI(2)(g), ...

`F_(2)(g) +2HCI(g) rarr 2HF(g) +CI_(2)(g), DeltaH^(Theta) =- 352.18 kJ`
Given heat of formation of `HF, Delta_(f)H^(Theta)underset((HF))" =- 268.3 kJ` The heat of formation of `HCl` will be

A

`-22 kJ mol^(-1)`

B

`88 kJ mol^(-1)`

C

`-92.21 kJ mol^(-1)`

D

`-183.8 kJ mol^(-1)`

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To find the heat of formation of HCl (Δ_fH°(HCl)), we can use the given reaction and the heat of formation of HF. Here’s a step-by-step solution: ### Step 1: Write down the reaction and the given data The reaction is: \[ F_2(g) + 2HCl(g) \rightarrow 2HF(g) + Cl_2(g) \] Given: - ΔH° (reaction) = -352.18 kJ - Δ_fH° (HF) = -268.3 kJ ...
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