The enthalpy change (Delta H) for the reactio | vedclass

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

Question

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \De

Vedclass · Accepted Answer

$-87.42$

Vedclass · Answer

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.Rearranging for $\Delta U$: $\Delta U = \Delta H - \Delta n_g RT$.For the reaction $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$,the change in the number of gaseous moles is $\Delta n_g = n_p - n_r = 2 - (1 + 3) = -2$.Given $\Delta H = -92.38 \ kJ$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.Substituting the values: $\Delta U = -92.38 - (-2 	imes 8.314 	imes 10^{-3} 	imes 298)$.$\Delta U = -92.38 - (-4.957) = -92.38 + 4.957 = -87.423 \ kJ$.Thus,$\Delta U \approx -87.42 \ kJ$.

The enthalpy change $(\Delta H)$ for the reaction,$N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is $-92.38 \ kJ$ at $298 \ K$. The internal energy change $\Delta U$ at $298 \ K$ is .............. $kJ$

Similar Questions

Two mole of an ideal gas is expanded isothermally and reversibly from $1 \ L$ to $10 \ L$ at $300 \ K$. The enthalpy change (in $kJ$) for the process is .....$kJ$

Which expression is correct for the work done in adiabatic reversible expansion of an ideal gas?

Which of the following is true for an adiabatic process?

What is calorimetry?

Vedclass Test Series

Exam Paper Generator

Online Exam Module