Enthalpy is an extensive property. In general,if enthalpy of an overall reaction $A \to B$ along one route is $\Delta_r H$ and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ represent enthalpies of intermediate reactions leading to product $B$. What will be the relation between $\Delta_r H$ for overall reaction and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ etc. for intermediate reactions?
- A$\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \dots$
- B$\Delta_r H = \Delta_r H_1 - \Delta_r H_2 - \Delta_r H_3 - \dots$
- C$\Delta_r H = \Delta_r H_1 \times \Delta_r H_2 \times \Delta_r H_3 \times \dots$
- D$\Delta_r H = \Delta_r H_1 / \Delta_r H_2 / \Delta_r H_3 / \dots$
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Similar Questions
From the following data,the enthalpy of vaporization of liquid water in $KJ \, mol^{-1}$ will be:
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l); \Delta H = -285.77 \, KJ \, mol^{-1}$
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(g); \Delta H = -241.84 \, KJ \, mol^{-1}$
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l); \Delta H = -285.77 \, KJ \, mol^{-1}$
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(g); \Delta H = -241.84 \, KJ \, mol^{-1}$
Medium
View SolutionCalculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$ given the following:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
The heats of hydrogenation for $3-$-methylbutene and $2-$-pentene are $-30\, kcal/mol$ and $-28\, kcal/mol$ respectively. The heats of combustion of $2-$-methylbutane and pentane are $-784\, kcal/mol$ and $-782\, kcal/mol$ respectively. All the values are given under standard conditions. Taking into account that combustion of both alkanes gives the same products,what is $\Delta H$ (in $kcal/mol$) for the following reaction under same conditions?
$3-$-methylbutene $\rightleftharpoons$ $2-$-pentene
$3-$-methylbutene $\rightleftharpoons$ $2-$-pentene
From Kirchhoff's equation,which factor affects the heat of reaction?
Easy
View SolutionThe heat of formation of $H_2O_{(l)}$ is $-68.0 \ kcal$. The heat of formation of $H_2O_{(g)}$ is likely to be........$kcal$.
Medium
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