For the complete combustion of ethanol, $$C2H5OHl+3O2g$$→$$2CO2g+3H2Ol$$ the amount of heat produced as measured in bomb calorimeter is $$1364.47$$ kJ $$mol-1$$ at $$250C$$. Assuming ideality, the enthalpy of combustion, ∆HC, for the reaction will $$beR=8.314$$ $$JK-1$$ $$mol-1$$

Question

For the complete combustion of ethanol, $$C2H5OHl+3O2g$$→$$2CO2g+3H2Ol$$ the amount of heat produced as measured in bomb calorimeter is $$1364.47$$ kJ $$mol-1$$ at $$250C$$. Assuming ideality, the enthalpy of combustion, ∆HC, for the reaction will $$beR=8.314$$ $$JK-1$$ $$mol-1$$

Infinity Learn · Accepted Answer

$$C2H5OHl+3O2g$$→$$2CO2g+3H2OlAmount$$ of heat produced in bomb calorimeter,∆$$U=-1364.47$$  kJ  $$mol-1Enthalpy$$ of a combustion reaction is∆$$H=$$∆$$U+$$∆ngRT where  ∆$$U=internal$$ energy ∆ng $$=$$ moles of gas products $$-$$ reactants $$R=Gas$$ constant $$T=Temperature$$ in K As per equation, ∆ng $$=2$$ $$-$$ $$3$$ $$=$$ $$-1$$ T $$=$$ $$250C=$$ $$25+273$$ K ⇒ $$T=298$$ K Thus,     ∆$$H=$$ $$-1364.47$$ $$+$$ $$-1$$ × $$8.314$$ × $$2981000$$ ∆$$H=$$ $$-1364.47$$ $$-$$ $$2.477$$ $$=$$ $$-1366.647$$ kJ $$mol-1$$ Hence, the enthalpy of combustion, ∆HC, for the given reaction will be $$-1366.947$$ kJ $$mol-1$$.

For the complete combustion of ethanol, C2H5OHl+3O2g→2CO2g+3H2Ol the amount of heat produced as measured in bomb calorimeter is 1364.47 kJ mol-1 at 250C. Assuming ideality, the enthalpy of combustion, ∆HC, for the reaction will beR=8.314 JK-1 mol-1

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