ThermochemistryMedium

Question

The standard enthalpies of formation of H2O(l), Li+ (aq) and OH− (aq) are −285.8, −278.5 and −228.9 kJ/mol, respectively. The standard enthalpy change for the below reaction is

2Li(s) + 2H2O(l) → 2Li+ (aq) + 2OH− (aq) + H2(g)

Options

A.+443.2 kJ
B.−443.2 kJ
C.−221.6 kJ
D.+221.6 kJ

Solution

$\Delta_{r}H = \sum_{}^{}{\Delta_{f}H_{\text{Products}}} - \sum_{}^{}{\Delta_{f}H_{\text{Reactants}}}$

$= \left\lbrack 2 \times \Delta_{f}H_{Li^{+}(aq)} + 2 \times \Delta_{f}H_{OH^{-}(aq)} + \Delta_{f}H_{H_{2}(g)} \right\rbrack - \left\lbrack 2 \times \Delta_{f}H_{Li(s)} + 2 \times \Delta_{f}H_{H_{2}O(l)} \right\rbrack $$$= \left\lbrack 2 \times ( - 278.5) + 2 \times ( - 228.9) + 0 \right\rbrack - \left\lbrack 2 \times 0 + 2 \times ( - 285.8) \right\rbrack = - 443.2\text{ KJ}$$

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