For the reaction N_2(g) + 3H_2(g) rightarrow | vedclass

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

Question

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

Vedclass · Accepted Answer

$-87.43$

Vedclass · Answer

(B) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$. For the reaction $N_2(g) + 3H_2(g) ightarrow 2NH_3(g)$,the change in the number of gaseous moles is $\Delta n_g = 2 - (1 + 3) = -2$. Given: $\Delta H = -92.38 \ kJ$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$. Substituting the values: $-92.38 = \Delta U + (-2) 	imes (8.314 	imes 10^{-3}) 	imes 298$. $-92.38 = \Delta U - 4.955$. $\Delta U = -92.38 + 4.955 = -87.425 \ kJ \approx -87.43 \ kJ$.

For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ at $298 \ K$,the enthalpy change $\Delta H$ is $-92.38 \ kJ$. What is the value of $\Delta U$ in $kJ$ at $298 \ K$?

Similar Questions

For an isochoric process,the first law of thermodynamics can be expressed as:

Internal energy of an ideal gas increases in:

$5 \, mol$ of an ideal gas expands isothermally and irreversibly from a pressure of $10 \, atm$ to $1 \, atm$ against a constant external pressure of $1 \, atm$. The work done $(w_{irr})$ at $300 \, K$ is:

Which among the following reactions exhibits $\Delta H = \Delta U$?

In an isochoric process,the increase in internal energy is

Vedclass Test Series

Exam Paper Generator

Online Exam Module