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A schematic plot of 1n K_eq versus inverse of temperature for a reaction is shown in the figure. The reaction must be

  • Option 1)

    exothermic

  • Option 2)

    endothermic

  • Option 3)

    one with negligible enthalpy change

  • Option 4)

    highly spontaneous at ordinary temperature

 

Answers (1)

best_answer

As we have learnt,

 

Arrhenius Equation -

Arrhenius gave the quantitative dependence of rate constant on temperature by the Arrhenius equation.

- wherein

k = A e^{-E_{a}/RT}

In k=InA-\frac{E_{a}}{RT}

k = Rate constant

 

 From the graph,

\\*\ln\left(\frac{K_2}{K_1} \right ) = \frac{\Delta H}{R}\left[\frac{1}{T_1} - \frac{1}{T_2} \right ] \\* \ln \left(\frac{6}{2} \right ) = \frac{\Delta H}{R}(-0.5\times 10^{-3})

\Rightarrow \Delta H of reaction comes out to be negetive. Hence the reaction is exothermic reaction.

 


Option 1)

exothermic

Option 2)

endothermic

Option 3)

one with negligible enthalpy change

Option 4)

highly spontaneous at ordinary temperature

Posted by

SudhirSol

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