Answer to Question #87420 in General Chemistry for jason

Question #87420
What is the standard enthalpy change for vaporizing 4.000 mol C5H12O(L) at 25°C?

ΔH° = ________ kJ
1
Expert's answer
2019-04-02T05:57:34-0400
C5H12O(l)C5H12O(g)C_5H_{12}O(l) \rightarrow C_5H_{12}O(g)

ΔHf0(C5H12O(l))=351.6kJmol\Delta H_f^0(C_5H_{12}O(l)) = -351.6 \frac{kJ}{mol}


ΔHf0(C5H12O(g))=294.6kJmol\Delta H_f^0 (C_5H_{12}O(g)) = -294.6 \frac{kJ}{mol}



ΔHvap=ΔHproductsΔHreactants\Delta H_{vap} = \sum \Delta H_{products} - \sum \Delta H_{reactants}

ΔHvap=ΔHf0(C5H12O(g))ΔHf0(C5H12O(l))=294.6kJmol(351.6kJmol)=57kJmol\Delta H_{vap} = \Delta H_f^0 (C_5H_{12}O(g))-\Delta H_f^0(C_5H_{12}O(l))= -294.6 \frac{kJ}{mol} -(-351.6 \frac{kJ}{mol}) = 57 \frac{kJ}{mol}

ΔH0=ΔHvap×n=57kJmol×4.000mol=228.0kJ\Delta H^0 = \Delta H_{vap} \times n = 57\frac{kJ}{mol}\times 4.000 mol = 228.0 kJ


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