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Current Question (ID: 8258)
Question:
$\text{Assume each reaction is carried out in an open container. For which of the following reactions will } \Delta H \text{ be equal to } \Delta U?$
Options:
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1. $\text{PCl}_5(\text{g}) \rightarrow \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g})$
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2. $2\text{CO}(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g})$
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3. $\text{H}_2(\text{g}) + \text{Br}_2(\text{g}) \rightarrow 2\text{HBr}(\text{g})$
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4. $\text{C}(\text{s}) + 2\text{H}_2\text{O}(\text{g}) \rightarrow 2\text{H}_2(\text{g}) + \text{CO}_2(\text{g})$
Solution:
\textbf{Hint:} \text{Change in number of moles of gases should be Zero}
\Delta H = \Delta U + \Delta n_g RT
\Delta n = 0
\Delta H = \Delta U
\Delta n_g = \sum \text{Gaseous P} - \sum \text{Gas R}
\text{Let us calculate } \Delta n_g \text{ for each reaction:}
\textbf{1. } \text{PCl}_5(\text{g}) \to \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g})
\Delta n_g = (1 + 1) - 1 = +1
\textbf{2. } 2\text{CO}(\text{g}) + \text{O}_2(\text{g}) \to 2\text{CO}_2(\text{g})
\Delta n_g = 2 - (2 + 1) = -1
\textbf{3. } \text{H}_2(\text{g}) + \text{Br}_2(\text{g}) \to 2\text{HBr}(\text{g})
\Delta n_g = 2 - (1 + 1) = 0
\textbf{4. } \text{C}(\text{s}) + 2\text{H}_2\text{O}(\text{g}) \to 2\text{H}_2(\text{g}) + \text{CO}_2(\text{g})
\Delta n_g = (2 + 1) - 2 = +1
\text{For } \Delta H = \Delta U, \text{ we need } \Delta n_g = 0
\text{Only reaction 3 has } \Delta n_g = 0, \text{ therefore only for this reaction } \Delta H = \Delta U.
\text{The relationship between enthalpy and internal energy is:}
\Delta H = \Delta U + \Delta n_g RT
\text{When } \Delta n_g = 0 \text{ (no change in number of gaseous moles), the } \Delta n_g RT \text{ term becomes zero, making } \Delta H = \Delta U.
\text{Therefore, option (3) is correct.}
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