The combustion of benzene (l) gives CO2(g) and H2O(l) . Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25° C ; heat of combustion (in kJ mol-1 ) of benzene at constant pressure will be - (R = 8.314 JK-1 mol-1)

Question

The combustion of benzene (l) gives CO2(g) and H2O(l) . Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25° C ; heat of combustion (in kJ mol-1 ) of benzene at constant pressure will be - (R = 8.314 JK-1 mol-1)  (A) 4152.6 (B) -452.46 (C) 3260 (D) -3267.6

Tardigrade Admin · Accepted Answer

C 6 H 6( I )+(15/2) O 2( g ) longrightarrow 6 CO 2( g )+3 H 2 O ( I ) Δ n g =6-(15/2)=-(3/2) Δ H =Δ U +Δ n g RT =-3263.9+(-(3/2)) × 8.314 × 298 × 10-3 =-3263.9+(-3.71) =-3267.6 kJ mol -1

Q. The combustion of benzene (l) gives CO2​(g) and H2​O(l) . Given that heat of combustion of benzene at constant volume is −3263.9kJmol−1 at 25∘C ; heat of combustion (in kJmol−1 ) of benzene at constant pressure will be -(R=8.314JK−1mol−1)

4152.6

-452.46

3260

-3267.6

C6​H6​(I)+215​O2​(g)⟶6CO2​(g)+3H2​O(I) Δng​=6−215​=−23​ ΔH=ΔU+Δng​RT =−3263.9+(−23​)×8.314×298×10−3 =−3263.9+(−3.71) =−3267.6kJmol−1

$C_{6} H_{6} (I) + \frac{15}{2} O_{2} (g) ⟶ 6 C O_{2} (g) + 3 H_{2} O (I)$

$Δ n_{g} = 6 - \frac{15}{2} = - \frac{3}{2}$

$Δ H = Δ U + Δ n_{g} RT$

$= - 3263.9 + (- \frac{3}{2}) \times 8.314 \times 298 \times 1 0^{- 3}$

$= - 3263.9 + (- 3.71)$

$= - 3267.6 k J m o l^{- 1}$