Answer to Question #99373 in General Chemistry for Kdkf

Question

The atmosphere contains the highly reactive molecule, OH, which acts to remove selected pollutants. Use the values for \Delta H rxn given below to find the \Delta H rxn for the formation of OH and H from water.
1/2 H2(g) + 1/2 O2(g) \rightarrow OH(g) \DeltaHrxn = 42.1 kJ
H2(g)\rightarrow 2H(g) \Delta Hrxn = 435.9 kJ
H2(g) + 1/2 O2(g)\rightarrow H2O(g) \DeltaHrxn = -241.8 kJ
H2O(g) \rightarrow H(g) + OH(g) \DeltaHrxn = ?

Answer 1

(I) $\frac{1}{2}H_2(g)+ \frac{1}{2}O_2(g) \rightarrow OH(g)$ $\Delta H_{rxn} = 42.1 kJ$

(II) $H_2(g) \rightarrow 2H(g)$ $\Delta H_{rxn} = 435.9 kJ$

(III) $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$ $\Delta H_{rxn} = -241.8 kJ$

To get $\Delta H_{rxn}$ for the formation of OH and H from water we should add $- (III) + (I)+\frac{(II)}{2}$

-(III) $H_2O(g) \rightarrow H_2(g) + \frac{1}{2}O_2(g)$ $\Delta H_{rxn} = 241.8 kJ$

(I) $\frac{1}{2}H_2(g) + \frac{1}{2} O_2(g)\rightarrow OH(g)$ $\Delta H_{rxn} = 42.1 kJ$

$\frac{(II)}{2}$ $\frac {1}{2}H_2 (g) \rightarrow H(g)$ $\Delta H_{rxn} = \frac{435.9}{2} = 217.95 kJ$

After addition:

$H_2O(g) + \frac{1}{2}H_2(g)+\frac{1}{2}O_2(g) + \frac{1}{2}H_2(g) \rightarrow H_2(g)+\frac{1}{2}O_2(g) + OH(g)+ H(g)$

$H_2O(g) \rightarrow OH(g) + H(g)$ $\Delta H_{rxn} = 241.8 + 42.1 +217.95 =501.85 kJ$