The molar enthalpy of vaporization of C 6 H 6 at its boiling point (353 K ) is 7.4 k Cal / mol. The molar internal energy change of vaporization is:

Question

The molar enthalpy of vaporization of C 6 H 6 at its boiling point (353 K ) is 7.4 k Cal / mol. The molar internal energy change of vaporization is: (A) 7.4 kCal / mol (B) 8.106 kCal / mol (C) 62.47 kCal / mol (D) 6.694 kCal / mol

Tardigrade Admin · Accepted Answer

C 6 H 6(l) leftharpoons C 6 H 6( g ) Δ H =Δ E +Δ n g RT 7.4=Δ E +1 × 2 × 10-3 × 353 7.4=Δ E +0.706 7.4-0.706=Δ E 6.694 k Cal / mol =Δ E

Q. The molar enthalpy of vaporization of $C_{6} H_{6}$ at its boiling point $(353 K)$ is $7.4 k C a l / m o l$ . The molar internal energy change of vaporization is:

7.4 kCal / mol

8.106 kCal / mol

62.47 kCal / mol

6.694 kCal / mol

$C_{6} H_{6} (l) ⇌ C_{6} H_{6} (g)$

$Δ H = Δ E + Δ n_{g} RT$

$7.4 = Δ E + 1 \times 2 \times 1 0^{- 3} \times 353$

$7.4 = Δ E + 0.706$

$7.4 - 0.706 = Δ E$

$6.694 k C a l / m o l = Δ E$

Q. The molar enthalpy of vaporization of C6​H6​ at its boiling point (353K) is 7.4kCal/mol. The molar internal energy change of vaporization is: