(B) For a reaction $X \to Y$,the enthalpy change $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_{a(f)

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(B) For a reaction $X \to Y$,the enthalpy change $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_{a(f)})$ and the activation energy of the reverse reaction $(E_{a(b)})$.$\Delta H = E_{a(f)} - E_{a(b)}$Given:$E_{a(f)} = 150\,kJ\,mol^{-1}$$\Delta H = -135\,kJ\,mol^{-1}$Substituting the values into the equation:$-135 = 150 - E_{a(b)}$Rearranging to solve for $E_{a(b)}$:$E_{a(b)} = 150 + 135 = 285\,kJ\,mol^{-1}$

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

Difficult

The reaction $X \to Y$ is an exothermic reaction. The activation energy of the forward reaction $X \to Y$ is $150\,kJ\,mol^{-1}$. The enthalpy of the reaction is $-135\,kJ\,mol^{-1}$. The activation energy for the reverse reaction,$Y \to X$,will be $.......\,kJ\,mol^{-1}$.

JEE MAIN 2013 All JEE MAIN

A
$280$
B
$285$
C
$270$
D
$15$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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According to the collision theory of chemical reactions:

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In a first order reaction at $27\,^oC$ and $47\,^oC$,$50\%$ of the reaction is complete in $30\,\min$ and $10\,\min$ respectively. Calculate the energy of activation $(E_a)$.

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Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

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Write the Arrhenius equation to calculate the activation energy $(E_a)$ at two different temperatures.

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When the temperature of a reaction is increased from $300 \ K$ to $310 \ K$,the rate of the reaction increases by $2.5$ times. If the rate constant of the reaction at $300 \ K$ is $K$,what will be the rate constant at $310 \ K$?

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The reaction $X \to Y$ is an exothermic reaction. The activation energy of the forward reaction $X \to Y$ is $150\,kJ\,mol^{-1}$. The enthalpy of the reaction is $-135\,kJ\,mol^{-1}$. The activation energy for the reverse reaction,$Y \to X$,will be $.......\,kJ\,mol^{-1}$.

Similar Questions

According to the collision theory of chemical reactions:

In a first order reaction at $27\,^oC$ and $47\,^oC$,$50\%$ of the reaction is complete in $30\,\min$ and $10\,\min$ respectively. Calculate the energy of activation $(E_a)$.

Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:$A \xrightarrow{K_1} B$,activation energy: $Ea_1$$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

Write the Arrhenius equation to calculate the activation energy $(E_a)$ at two different temperatures.

When the temperature of a reaction is increased from $300 \ K$ to $310 \ K$,the rate of the reaction increases by $2.5$ times. If the rate constant of the reaction at $300 \ K$ is $K$,what will be the rate constant at $310 \ K$?

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Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$