Solveeit Logo
Login

Question

Chemistry Question on Thermodynamics terms

The combustion of benzene (l) (l) gives CO2(g)CO_2(g) and H2O(l)H_2O(l) . Given that heat of combustion of benzene at constant volume is −3263.9 kJ mol−1-3263.9 \, kJ \, mol^{-1} at 25∘C25^{\circ} C ; heat of combustion (in kJ mol−1 kJ \, mol^{-1} ) of benzene at constant pressure will be -(R=8.314 JK−1 mol−1)(R = 8.314 \, JK^{-1} \, mol^{-1})

A

4152.6

B

-452.46

C

3260

D

-3267.6

Answer

-3267.6

Explanation

Solution

C6H6(I)+152O2(g)⟶6CO2(g)+3H2O(I)C _{6} H _{6}( I )+\frac{15}{2} O _{2}( g ) \longrightarrow 6 CO _{2}( g )+3 H _{2} O ( I ) Δng=6−152=−32\Delta n _{ g }=6-\frac{15}{2}=-\frac{3}{2} ΔH=ΔU+ΔngRT\Delta H =\Delta U +\Delta n _{ g } RT =−3263.9+(−32)×8.314×298×10−3=-3263.9+\left(-\frac{3}{2}\right) \times 8.314 \times 298 \times 10^{-3} =−3263.9+(−3.71)=-3263.9+(-3.71) =−3267.6 kJ mol−1=-3267.6 \,kJ\, mol ^{-1}