(B) The relationship between the activation energy of the forward reaction $(E_a)_f$,the backward reaction $(E_a)_b$,and the enthalpy change $\Delta H

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(B) The relationship between the activation energy of the forward reaction $(E_a)_f$,the backward reaction $(E_a)_b$,and the enthalpy change $\Delta H$ is given by: $(E_a)_f - (E_a)_b = \Delta H$. Given that the reaction is endothermic,$\Delta H = +30 \ kcal$. Given $(E_a)_b = 50 \ kcal$. Substituting these values into the equation: $(E_a)_f - 50 \ kcal = 30 \ kcal$. Therefore,$(E_a)_f = 30 \ kcal + 50 \ kcal = 80 \ kcal$.

Class 11 Chemistry Thermodynamics Heat of reaction, Bond energy and Hess law

Medium

For a reaction $A + B + 30 \ kcal \longrightarrow \text{Product}$,the energy of activation of its backward reaction is $50 \ kcal$. The energy of activation of its forward reaction is :-

A
Equal to $50 \ kcal$
B
Greater than $50 \ kcal$
C
Less than $50 \ kcal$
D
Either greater or less than $50 \ kcal$

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Heat of reaction, Bond energy and Hess law

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For a reaction $A + B + 30 \ kcal \longrightarrow \text{Product}$,the energy of activation of its backward reaction is $50 \ kcal$. The energy of activation of its forward reaction is :-

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The heat released when $0.8 \, g$ of carbon is converted to carbon dioxide is $x \, cal$. The heat released when $0.8 \, g$ of carbon is converted to carbon monoxide is $y \, cal$. If $x > y$,then the heat released when $1.86 \, g$ of carbon monoxide is converted to carbon dioxide will be:

Given below are two statements:Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.The correct answer is:

The standard enthalpies of formation of $CO_{2(g)}$,$H_2O_{(\ell)}$ and glucose$_{(s)}$ at $25^{\circ} C$ are $-400 \ kJ/mol$,$-300 \ kJ/mol$ and $-1300 \ kJ/mol$,respectively. The standard enthalpy of combustion per gram of glucose at $25^{\circ} C$ is

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The $\Delta H_f^o$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is

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Given below are two statements:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is: