The enthalpy of a reaction at 273 , K is -3.5 | vedclass

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Question

(A) According to Kirchhoff's law,the variation of enthalpy of reaction with temperature is given by: $\Delta H_2 - \Delta H_1 = \int_{T_1}^{T_2} \Delt

Vedclass · Accepted Answer

$-3.57 \, kJ$

Vedclass · Answer

(A) According to Kirchhoff's law,the variation of enthalpy of reaction with temperature is given by: $\Delta H_2 - \Delta H_1 = \int_{T_1}^{T_2} \Delta C_p \, dT$. Given that $\Delta C_p = 0$,the equation becomes $\Delta H_2 - \Delta H_1 = 0$,which implies $\Delta H_2 = \Delta H_1$. Therefore,the enthalpy of the reaction remains constant at all temperatures when $\Delta C_p = 0$. Thus,the enthalpy at $373 \, K$ is the same as at $273 \, K$,which is $-3.57 \, kJ$.

The enthalpy of a reaction at $273 \, K$ is $-3.57 \, kJ$. If $\Delta C_p = 0$,what will be the enthalpy of the reaction at $373 \, K$?

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