ThermochemistryHard

Question

Given enthalpy of formation of CO2(g) and CaO(s) are −94.0 kJ and −152 kJ, respectively, and the enthalpy of the reaction CaCO3(s) → CaO(s) + CO2(g) is 42 kJ. The enthalpy of formation of CaCO3(s) is

Options

A.−42 kJ/mol
B.−202 kJ/mol
C.+202 kJ/mol
D.−288 kJ/mol

Solution

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

Or, $42 = \left\lbrack ( - 152) + ( - 74) \right\rbrack - \left\lbrack \Delta_{f}H_{CaCO_{3}(s)} \right\rbrack$

$\therefore\Delta_{f}H_{CaCO_{3}(s)} = - 288\text{ kJ/mol}$

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