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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $373 \, K$?

Similar Questions

Which among the following pair of properties is intensive?

If $0.75 \ mol$ of an ideal gas expands isothermally at $27 \ ^{\circ}C$ from $15 \ L$ to $25 \ L$,the maximum work obtained is ...... $J$. (in $.5$)

Find the work done on $2 \ mol$ of an ideal gas at $27^{\circ} C$ if it is compressed reversibly and isothermally from $5.05 \times 10^6 \ Nm^{-2}$ to $1.01 \times 10^5 \ Nm^{-2}$ pressure. (in $kJ$)

If the work done during the isothermal reversible expansion of an ideal gas at a pressure of $10 \ atm$ from $4 \ L$ to a final volume is $-184.24 \ L \ atm$,the final volume of the gas in $L$ is

What are extensive properties?

Vedclass Test Series

Exam Paper Generator

Online Exam Module