QuestionAugust 15, 2025

Consider the chemical equations shown-here. NO(g)+O_(3)(g)arrow NO_(2)(g)+O_(2)(g)Delta H_(1)=-198.9kJ (3)/(2)O_(2)(g)arrow O_(3)(g)Delta H_(2)=142.3kJ O(g)arrow (1)/(2)O_(2)(g)Delta H_(3)=-247.5kJ What is Delta H_(rxn) for the reaction shown below? NO(g)+O(g)arrow NO_(2)(g) square

Consider the chemical equations shown-here. NO(g)+O_(3)(g)arrow NO_(2)(g)+O_(2)(g)Delta H_(1)=-198.9kJ (3)/(2)O_(2)(g)arrow O_(3)(g)Delta H_(2)=142.3kJ O(g)arrow (1)/(2)O_(2)(g)Delta H_(3)=-247.5kJ What is Delta H_(rxn) for the reaction shown below? NO(g)+O(g)arrow NO_(2)(g) square
Consider the chemical equations shown-here. NO(g)+O_(3)(g)arrow NO_(2)(g)+O_(2)(g)Delta H_(1)=-198.9kJ (3)/(2)O_(2)(g)arrow O_(3)(g)Delta H_(2)=142.3kJ O(g)arrow (1)/(2)O_(2)(g)Delta H_(3)=-247.5kJ What is Delta H_(rxn) for the reaction shown below? NO(g)+O(g)arrow NO_(2)(g) square

Solution
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Answer

\Delta H_{rxn} = -93.7 kJ Explanation 1. Write the target reaction NO(g) + O(g) \rightarrow NO_{2}(g) 2. Use Hess's Law Combine given reactions to form the target reaction using Hess's Law. 3. Reverse and adjust equations Reverse the second equation: O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g), \Delta H = -142.3 kJ. Reverse the third equation: \frac{1}{2}O_{2}(g) \rightarrow O(g), \Delta H = 247.5 kJ. 4. Add adjusted equations Add the first equation with reversed second and third equations: NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g), \Delta H = -198.9 kJ O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g), \Delta H = -142.3 kJ \frac{1}{2}O_{2}(g) \rightarrow O(g), \Delta H = 247.5 kJ 5. Calculate \Delta H_{rxn} Sum the enthalpies: \Delta H_{rxn} = (-198.9) + (-142.3) + 247.5 kJ

Explanation

1. Write the target reaction<br /> $NO(g) + O(g) \rightarrow NO_{2}(g)$<br /><br />2. Use Hess's Law<br /> Combine given reactions to form the target reaction using Hess's Law.<br /><br />3. Reverse and adjust equations<br /> Reverse the second equation: $O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g)$, $\Delta H = -142.3$ kJ.<br /> Reverse the third equation: $\frac{1}{2}O_{2}(g) \rightarrow O(g)$, $\Delta H = 247.5$ kJ.<br /><br />4. Add adjusted equations<br /> Add the first equation with reversed second and third equations:<br /> $NO(g) + O_{3}(g) \rightarrow NO_{2}(g) + O_{2}(g)$, $\Delta H = -198.9$ kJ<br /> $O_{3}(g) \rightarrow \frac{3}{2}O_{2}(g)$, $\Delta H = -142.3$ kJ<br /> $\frac{1}{2}O_{2}(g) \rightarrow O(g)$, $\Delta H = 247.5$ kJ<br /><br />5. Calculate $\Delta H_{rxn}$<br /> Sum the enthalpies: $\Delta H_{rxn} = (-198.9) + (-142.3) + 247.5$ kJ
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