The enthalpy of combustion of 2 moles of benzene at 27° C differs from the value determined in bomb calorimeter by

Question

The enthalpy of combustion of 2 moles of benzene at 27° C differs from the value determined in bomb calorimeter by (A) - 2.494 kJ (B) 2.494 kJ (C) - 7.483 kJ (D) 7.483 kJ

Tardigrade Admin · Accepted Answer

By bomb calorimeter we get Δ E. 2 C 6 H 6( l )+15 O 2( g ) longrightarrow 12 CO 2( g )+6 H 2 O (l) Δ H -Δ E =Δ nRT =(12-15) × 8.314 × 300=-7.483 kJ

Q. The enthalpy of combustion of $2$ moles of benzene at $2 7^{\circ} C$ differs from the value determined in bomb calorimeter by

- 2.494 kJ

2.494 kJ

- 7.483 kJ

7.483 kJ

By bomb calorimeter we get $Δ E$ .
$2 C_{6} H_{6} (l) + 15 O_{2} (g) ⟶ 12 C O_{2} (g) + 6 H_{2} O (l)$
$Δ H - Δ E = Δ n RT$
$= (12 - 15) \times 8.314 \times 300 = - 7.483 k J$

Q. The enthalpy of combustion of 2 moles of benzene at 27∘C differs from the value determined in bomb calorimeter by