At 25°C, the enthalpy of the following processes is given:H2 (g)+O2(g)→2OH(g) ; ΔH0=78kjmol−1H2(g)+1/2O2(g)→H2O(g) ; ΔH0=−242kjmol−1H2(g)→2H(g) ;ΔH0=436kjmol−1H2O(g) → H(g)+OH(g) ; ΔH0=X kJ mol−1.Find the value of X−19950

H2O(g)→H(g)+OH(g) ΔH0=kJmol−1 2H2O 2H2+O2 ΔH=+242×2=484 H2+O2→2OH ΔH=78 2H2O→H2+2OH ΔH=562 H2→2H ΔH=436 2H2O→2H+2 OH ΔH=998 H2O→H+OH ΔH=499=X X−19950=6

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At 25°C, the enthalpy of the following processes is given:
$\begin{array}{l} H_{2}_{(g)} + {O_{2}}_{(g)} \to 2 {OH}_{(g)}; {ΔH}^{0} = 78 kj {mol}^{- 1} \\ {H_{2}}_{(g)} + 1 / 2 {O_{2}}_{(g)} \to H_{2} O_{(g)}; {ΔH}^{0} = - 242 kj {mol}^{- 1} \\ {H_{2}}_{(g)} \to 2 H_{(g)}; {ΔH}^{0} = 436 kj {mol}^{- 1} \end{array}$
$H_{2} O_{(g)} \to H_{(g)} + {OH}_{(g)}; {ΔH}^{0} = X kJ {mol}^{- 1}$ .
Find the value of $\frac{X - 199}{50}$

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answer is 6.

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Detailed Solution

$H_{2} O (g) \to H (g) + OH (g) {ΔH}^{0} = kJ {mol}^{- 1} 2 H_{2} O 2 H_{2} + O_{2} ΔH = + 242 \times 2 = 484 H_{2} + O_{2} \to 2 OH ΔH = 78 2 H_{2} O \to H_{2} + 2 OH ΔH = 562 H_{2} \to 2 H ΔH = 436 2 H_{2} O \to 2 H + 2 O H ΔH = 998 H_{2} O \to H + OH ΔH = 499 = X \frac{X - 199}{50} = 6$