For the reaction,$C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,$\Delta U$ is the heat of reaction at constant volume. Then the heat of reaction at constant pressure is:
- A$\Delta H = \Delta U + RT$
- B$\Delta H = \Delta U + 2RT$
- C$\Delta H = \Delta U - 2RT$
- D$\Delta H = \Delta U - RT$
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Similar Questions
Calculate the $N-N$ bond energy in $N_2H_4$ from the given bond enthalpy data.
$\varepsilon_{N-H} = 393 \ kJ/mol$
$\varepsilon_{H-H} = 436 \ kJ/mol$
$\Delta H_{vap}[N_2H_{4(l)}] = 18 \ kJ/mol$
$N_2H_{4(l)} + H_{2(g)} \to 2NH_{3(g)} : \Delta H = -142 \ kJ/mol$
....... $kJ/mol$
$\varepsilon_{N-H} = 393 \ kJ/mol$
$\varepsilon_{H-H} = 436 \ kJ/mol$
$\Delta H_{vap}[N_2H_{4(l)}] = 18 \ kJ/mol$
$N_2H_{4(l)} + H_{2(g)} \to 2NH_{3(g)} : \Delta H = -142 \ kJ/mol$
....... $kJ/mol$
Medium
View SolutionIf the standard enthalpy change $\left(\Delta_{r} H^\theta\right)$ for the reaction $H_{2(g)} + Br_{2(l)} \rightarrow 2 HBr_{(g)}$ is $-72.8 \ kJ$,the standard enthalpy of formation $\left(\Delta_{f} H^\theta\right)$ of $HBr_{(g)}$ (in $kJ \ mol^{-1}$) is
Considering the reaction $3O_2 \rightarrow 2O_3$; $\Delta H = +ve$,what can we conclude about the reaction?
Medium
View SolutionCalculate $\Delta H$ for the reaction: $H_{2(g)} + O_{2(g)} \rightarrow H_2O_{2(g)}$ given the bond energies: $BE_{H-H} = 436 \ kJ/mol$,$BE_{O=O} = 499 \ kJ/mol$,$BE_{O-O} = 142 \ kJ/mol$,and $BE_{O-H} = 460 \ kJ/mol$. (in $kJ$)
Easy
View SolutionIf $H^{+} + OH^{-} \to H_2O + 13.7 \ kcal$,then the heat of neutralization for complete neutralization of one mole of $H_2SO_4$ by base will be.......$kcal$
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