QuestionMay 7, 2025

61. Given the following data 2O_(3)(g)arrow 3O_(2)(g) Delta H=-427kJ O_(2)(g)arrow 2O(g) Delta H=+495kJ NO(g)+O_(3)(g)arrow NO_(2)(g)+O_(2)(g) Delta H=-199kJ calculate Delta H for the reaction NO(g)+O(g)arrow NO_(2)(g)

61. Given the following data 2O_(3)(g)arrow 3O_(2)(g) Delta H=-427kJ O_(2)(g)arrow 2O(g) Delta H=+495kJ NO(g)+O_(3)(g)arrow NO_(2)(g)+O_(2)(g) Delta H=-199kJ calculate Delta H for the reaction NO(g)+O(g)arrow NO_(2)(g)
61. Given the following data 2O_(3)(g)arrow 3O_(2)(g) Delta H=-427kJ O_(2)(g)arrow 2O(g) Delta H=+495kJ NO(g)+O_(3)(g)arrow NO_(2)(g)+O_(2)(g) Delta H=-199kJ calculate Delta H for the reaction NO(g)+O(g)arrow NO_(2)(g)

Solution
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Answer

\Delta H = -107 \text{ kJ} Explanation 1. Write the target reaction The target reaction is NO(g) + O(g) \rightarrow NO_{2}(g). 2. Use Hess's Law According to Hess's Law, the enthalpy change for a reaction is the sum of the enthalpy changes for each step that leads to the overall reaction. 3. Rearrange given reactions We need to manipulate the given reactions to derive the target reaction: 1. Reverse the second reaction: 2O(g) \rightarrow O_{2}(g), \Delta H = -495 \text{ kJ}. 2. Divide the first reaction by 2: O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g), \Delta H = -213.5 \text{ kJ}. 3. Use the third reaction as it is: NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g), \Delta H = -199 \text{ kJ}. 4. Combine reactions Add the manipulated reactions: - From reaction 1 (reversed): O_{2}(g) \rightarrow 2O(g), \Delta H = +495 \text{ kJ}. - From reaction 2 (divided): O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g), \Delta H = -213.5 \text{ kJ}. - Reaction 3: NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g), \Delta H = -199 \text{ kJ}. Sum these reactions to get: NO(g) + O(g) \rightarrow NO_{2}(g) 5. Calculate total \Delta H Total \Delta H = (-495 \text{ kJ}) + (-213.5 \text{ kJ}) + (-199 \text{ kJ}) = -107 \text{ kJ}.

Explanation

1. Write the target reaction<br /> The target reaction is $NO(g) + O(g) \rightarrow NO_{2}(g)$.<br /><br />2. Use Hess's Law<br /> According to Hess's Law, the enthalpy change for a reaction is the sum of the enthalpy changes for each step that leads to the overall reaction.<br /><br />3. Rearrange given reactions<br /> We need to manipulate the given reactions to derive the target reaction:<br />1. Reverse the second reaction: $2O(g) \rightarrow O_{2}(g)$, $\Delta H = -495 \text{ kJ}$.<br />2. Divide the first reaction by 2: $O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g)$, $\Delta H = -213.5 \text{ kJ}$.<br />3. Use the third reaction as it is: $NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g)$, $\Delta H = -199 \text{ kJ}$.<br /><br />4. Combine reactions<br /> Add the manipulated reactions:<br />- From reaction 1 (reversed): $O_{2}(g) \rightarrow 2O(g)$, $\Delta H = +495 \text{ kJ}$.<br />- From reaction 2 (divided): $O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g)$, $\Delta H = -213.5 \text{ kJ}$.<br />- Reaction 3: $NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g)$, $\Delta H = -199 \text{ kJ}$.<br /><br /> Sum these reactions to get:<br />$NO(g) + O(g) \rightarrow NO_{2}(g)$<br /><br />5. Calculate total $\Delta H$<br /> Total $\Delta H = (-495 \text{ kJ}) + (-213.5 \text{ kJ}) + (-199 \text{ kJ}) = -107 \text{ kJ}$.
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