ThermochemistryHard
Question
Calculate Δf Ho for aqueous chloride ion from the following data.
½ H2(g) + ½ Cl2(g) → HCl(g): ΔfHo = −92.4 kJ
HCl(g) + nH2O(l) → H+ (aq) + Cl− (aq): ΔHo = −74.8 kJ
ΔfHo (H+, aq.) = 0.0 kJ
Options
A.0.0
B.+83.6 kJ
C.+167.2 kJ
D.−167.2 kJ
Solution
$(a)\frac{1}{2}H_{2}(g) + \frac{1}{2}Cl_{2}(g) \rightarrow HCl(g);\Delta H^{o} = - 92.4\text{ KJ}$
$(b)HCl(g) + nH_{2}O(l) \rightarrow H^{+}(aq) + Cl^{-}(aq);\Delta H^{o} = - 74.8\text{ kJ} $$${(c)\frac{1}{2}H_{2}(g) + nH_{2}O(l) \rightarrow H^{+}(aq);\Delta H^{o} = 0 }{\text{Required T.C.E. is} }{\frac{1}{2}Cl_{2}(g) + nH_{2}O(l) \rightarrow Cl^{-}(aq);\Delta H^{o} = ? }{\text{From }(a) + (b) - (c);\Delta H^{o} = - 167.2\text{ KJ}}$$
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