Question

The average O–H bond energy in H2O with the help of following data.

(1) H2O(ℓ) $\longrightarrow$H2O(g) ; ΔH = + 40.6 KJ mol–1

(2) 2H(g) $\longrightarrow$ H2 (g) ; ΔH = – 435.0 KJ mol–1

(3) O2(g) $\longrightarrow$ 2O(g) ; ΔH = + 489.6 KJ mol–1

(4) 2H2 (g) + O2 (g) $\longrightarrow$2H2O(ℓ) ; ΔH = – 571.6 KJ mol–1

• A. 584.9 KJ mol–1
• B. 279.8 KJ mol–1
• C. 462.5 KJ mol–1
• D. 925 KJ mol–1

Answer 1

Answer: (C)

462.5 KJ mol–1

Answer 2

(1) H2O($\mathcal{l}$)$\longrightarrow$H2O(g) $\Delta$H = 40.6 KJ/mole

(2) 2H(g) $\longrightarrow$ H2(g) $\Delta$H = –435.0 KJ/mole

(3) 2O(g) $\longrightarrow$ O2(g) $\Delta$H = –489.6 KJ/mole

(4)

$\Delta$H = –571.6 KJ/mole

(1) Calculation of $\Delta$H°f (H2O, $\mathcal{l}$)

2H2(g) + O2(g) ® 2H2O($\mathcal{l}$)$\Delta$H = – 571.6 KJ/mole

$\Delta$H°r = 2$\Delta$H°F(H2O, $\mathcal{l}$) – 2$\Delta$H°F{H2,(g)} – $\Delta$H°F(O2, g)

$\downarrow$ $\downarrow$

Zero Zero

–571.6 = 2$\Delta$H°F(H2O, $\mathcal{l}$) so $\Delta$H°F(H2O, $\mathcal{l}$) = – 285.8

(2) Calculation of $\Delta$H°F(H2O, g)

H2O($\mathcal{l}$)$\longrightarrow$ H2O(g) $\Delta$H = 40.6

$\Delta$Hr = $\Delta$H°F (H2O, g) – $\Delta$H° (H2O, $\mathcal{l}$)

$\Delta$H°F (H2O, g) = $\Delta$H°F (H2O, $\mathcal{l}$) + DHr

= –285.8 + 40 = –245.8

(3)

$\Delta$H = –245.8

$\Delta$Hr = $\in$H–H + $\frac{1}{2} \in$O–O – 2 $\in$O–H

$\Rightarrow$–245.8 = + 435 + $\frac{1}{2}$ (489.6) – 2 × $\in$O–H

2$\in$O–H = 435 + 244.8 + 245.8

$\Rightarrow$ 2$\in$O–H = 925.6

$\in$O–H = 462.8