Q. For the gaseous reaction involving the complete combustion of isobutane :

4411 241 BITSATBITSAT 2005 Report Error

A

$Δ H = Δ E$

11%

B

$Δ H > Δ E$

56%

C

$Δ H < Δ E$

30%

D

none of these

4%

Solution:

Equation of combustion of isobutane will be as follows:
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (g)$
$∵ Δ H = Δ E + Δ n RT$
where,
$Δ n =$ number of gaseous moles (products)
$Δ n = (4 + 5) - (1 + \frac{13}{2})$ number of moles of gaseous reactants.
$Δ n = + \frac{3}{2} (+ v e)$
Since, $Δ n$ is + ve
$Δ H$ must be greater than $Δ E$ .
$∴ Δ H > Δ E$