Consider the combustion of liquid benzene at 298 ,K: Data: Molar enthalpy of…

CHEMISTRY · Single correct · A Level · Chemistry · Energetics · Energetics

Problem

Consider the combustion of liquid benzene at 298\,K: C6H6(l)+152O2(g)→6CO2(g)+3H2O(g)\text{C}_6\text{H}_6(l)+\tfrac{15}{2}\text{O}_2(g)\rightarrow 6\text{CO}_2(g)+3\text{H}_2\text{O}(g) Data: ΔHf∘(C6H6(l))=+49.0 kJ mol−1\Delta H_f^{\circ}(\text{C}_6\text{H}_6(l)) = +49.0\,\text{kJ mol}^{-1} ΔHf∘(CO2(g))=−393.5 kJ mol−1\Delta H_f^{\circ}(\text{CO}_2(g)) = -393.5\,\text{kJ mol}^{-1} ΔHf∘(H2O(l))=−285.8 kJ mol−1\Delta H_f^{\circ}(\text{H}_2\text{O}(l)) = -285.8\,\text{kJ mol}^{-1} Molar enthalpy of vaporization of water at 298\,K: ΔHvap(H2O)=+44.0 kJ mol−1\Delta H_{\text{vap}}(\text{H}_2\text{O}) = +44.0\,\text{kJ mol}^{-1} Calculate the standard enthalpy change ΔH∘\Delta H^{\circ} for the combustion reaction above. (a) −3.27×103 kJ-3.27\times10^{3}\,\text{kJ} (b) −2.65×103 kJ-2.65\times10^{3}\,\text{kJ} (c) −3.19×103 kJ-3.19\times10^{3}\,\text{kJ} (d) −3.04×103 kJ-3.04\times10^{3}\,\text{kJ}

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