Question

The heat of formation of C2H5OH($\mathcal{l}$) is - 66 kcal/mole. The heat of combustion of CH3OCH3 (g) is – 348 kcal/mole. $\Delta$Hf for H2O and CO2 are -68 kcal/mole and -94 kcal/mole respectively. Then, the $\Delta$H for the isomerisation reaction C2H5OH ($\mathcal{l}$)$\longrightarrow$ CH3OCH3(g), and $\Delta$E for the same are at T = 25⁰C

• A. $\Delta$H = 18 kcal/mole, $\Delta$E = 17.301 kcal/mole
• B. $\Delta$H = 22 kcal/mole, $\Delta$E = 21.408 kcal/mole
• C. $\Delta$H = 26 kcal/mole, $\Delta$E = 25.709 kcal/mole
• D. $\Delta$H = 30 kcal/mole, $\Delta$E = 28.522 kcal/mole

Answer 1

Answer: (B)

$\Delta$H = 22 kcal/mole, $\Delta$E = 21.408 kcal/mole

Answer 2

$\Delta$HF⁰ C2H5OH ($\mathcal{l}$) = – 66 Kcal/mole

2C + 3H2$\frac{1}{2}$+ O2 $\longrightarrow$ C2H5OH

$\Delta$H = – 66 kcal/mole ....(1)

CH3 – O – CH3 + 3O2 $\longrightarrow$ 2 CO2 + 3H2O $\Delta$H = – 348 kcal/mole ....(2)

[H2(g) + $\frac{1}{2}$O2(g) $\longrightarrow$ H2O

$\Delta$H3 – 68 kcal/mole ....(3)

[C + O2 $\longrightarrow$ CO2

$\Delta$H4 = – 94 kcal/mole ....(4)

Target equation = – eq 1 – eq 2 + 3eq 3 + 2 eq 2

$\Delta$H = + 66 + 348 – 3 × 68 – 2 × 94 = + 66 + 348 – 204 – 188 $\Rightarrow$ $\Delta$H = 22 K cal/mole

$\Delta$H = $\Delta$E + $\Delta$ngRT

$\Rightarrow$ 22 = $\Delta$E + 1 × 2 × 298 × 10–3

$\Rightarrow$ $\Delta$E = 21.4 Kcal/mole