QuestionMarch 20, 2026

Given the following reactions and corresponding Delta H values, find the Delta H (to the nearest whole #) for the reaction: H_(2)SO_(4)(l)arrow SO_(3)(g)+H_(2)O(g) H_(2)S(g)+2O_(2)(g)arrow H_(2)SO_(4)(l) Delta H=-235kJ H_(2)S(g)+2O_(2)(g)arrow SO_(3)(g)+H_(2)O(l) Delta H=-207kJ H_(2)O(l)arrow H_(2)O(g) Delta H=44kJ square

Given the following reactions and corresponding Delta H values, find the Delta H (to the nearest whole #) for the reaction: H_(2)SO_(4)(l)arrow SO_(3)(g)+H_(2)O(g) H_(2)S(g)+2O_(2)(g)arrow H_(2)SO_(4)(l) Delta H=-235kJ H_(2)S(g)+2O_(2)(g)arrow SO_(3)(g)+H_(2)O(l) Delta H=-207kJ H_(2)O(l)arrow H_(2)O(g) Delta H=44kJ square
Given the following reactions and corresponding Delta H values, find the Delta H (to the nearest whole #) for the reaction: H_(2)SO_(4)(l)arrow SO_(3)(g)+H_(2)O(g) H_(2)S(g)+2O_(2)(g)arrow H_(2)SO_(4)(l) Delta H=-235kJ H_(2)S(g)+2O_(2)(g)arrow SO_(3)(g)+H_(2)O(l) Delta H=-207kJ H_(2)O(l)arrow H_(2)O(g) Delta H=44kJ square

Solution
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Answer

\Delta H = 16\ \text{kJ} Explanation 1. Identify target reaction We need H_{2}SO_{4}(l) \rightarrow SO_{3}(g) + H_{2}O(g). 2. Use given reactions to form target Reaction 1: H_{2}S + 2O_{2} \rightarrow H_{2}SO_{4}(l), \Delta H = -235 kJ Reaction 2: H_{2}S + 2O_{2} \rightarrow SO_{3}(g) + H_{2}O(l), \Delta H = -207 kJ. Subtract Reaction 2 from Reaction 1 to get: H_{2}SO_{4}(l) \rightarrow SO_{3}(g) + H_{2}O(l). 3. Calculate \Delta H for intermediate reaction \Delta H = (-235) - (-207) = -235 + 207 = -28 kJ. 4. Add water phase change H_{2}O(l) \rightarrow H_{2}O(g), \Delta H = 44 kJ. Total \Delta H = -28 + 44 = 16 kJ.

Explanation

1. Identify target reaction <br /> We need $H_{2}SO_{4}(l) \rightarrow SO_{3}(g) + H_{2}O(g)$.<br /><br />2. Use given reactions to form target <br /> Reaction 1: $H_{2}S + 2O_{2} \rightarrow H_{2}SO_{4}(l)$, $\Delta H = -235$ kJ <br /> Reaction 2: $H_{2}S + 2O_{2} \rightarrow SO_{3}(g) + H_{2}O(l)$, $\Delta H = -207$ kJ. <br />Subtract Reaction 2 from Reaction 1 to get: <br />$H_{2}SO_{4}(l) \rightarrow SO_{3}(g) + H_{2}O(l)$.<br /><br />3. Calculate $\Delta H$ for intermediate reaction <br /> $\Delta H = (-235) - (-207) = -235 + 207 = -28$ kJ.<br /><br />4. Add water phase change <br /> $H_{2}O(l) \rightarrow H_{2}O(g)$, $\Delta H = 44$ kJ. <br />Total $\Delta H = -28 + 44 = 16$ kJ.
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