Calculate the Delta_fH^{circ} of N_2H_{4(g)} | vedclass

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

Question

(B) The formation reaction for $N_2H_{4(g)}$ is:$N_{2(g)} + 2H_{2(g)} \rightarrow N_2H_{4(g)}$The enthalpy of formation $\Delta_fH^{\circ}$ is calcula

Vedclass · Accepted Answer

$62 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The formation reaction for $N_2H_{4(g)}$ is:$N_{2(g)} + 2H_{2(g)} ightarrow N_2H_{4(g)}$The enthalpy of formation $\Delta_fH^{\circ}$ is calculated using bond energies as:$\Delta_fH^{\circ} = \sum B.E.(reactants) - \sum B.E.(products)$$\Delta_fH^{\circ} = [B.E.(N \equiv N) + 2 	imes B.E.(H-H)] - [B.E.(N-N) + 4 	imes B.E.(N-H)]$Substituting the given values:$\Delta_fH^{\circ} = [941 + 2(436)] - [159 + 4(398)]$$\Delta_fH^{\circ} = [941 + 872] - [159 + 1592]$$\Delta_fH^{\circ} = 1813 - 1751$$\Delta_fH^{\circ} = 62 \ kJ \ mol^{-1}$

Calculate the $\Delta_fH^{\circ}$ of $N_2H_{4(g)}$ given the following bond energies:
$B.E.(N-N) = 159 \ kJ \ mol^{-1}$
$B.E.(H-H) = 436 \ kJ \ mol^{-1}$
$B.E.(N \equiv N) = 941 \ kJ \ mol^{-1}$
$B.E.(N-H) = 398 \ kJ \ mol^{-1}$

Similar Questions

The heat evolved per mole of $H^+$ ions during the neutralization of a strong acid and a strong base is ......

In the reaction $S + \frac{3}{2} O_{2} \rightarrow SO_{3} + 2x \ kJ$ and $SO_{2} + \frac{1}{2} O_{2} \rightarrow SO_{3} + y \ kJ$,the heat of formation of $SO_{2}$ is

The standard enthalpies of formation of $CO_{2(g)}$,$H_2O_{(\ell)}$ and glucose$_{(s)}$ at $25^{\circ} C$ are $-400 \ kJ/mol$,$-300 \ kJ/mol$ and $-1300 \ kJ/mol$,respectively. The standard enthalpy of combustion per gram of glucose at $25^{\circ} C$ is

Which is the best definition of “heat of neutralization”?

For the reaction $C(s) + O_2(g) \rightarrow CO_2(g)$,the enthalpy change $(\Delta H)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Calculate the $\Delta_fH^{\circ}$ of $N_2H_{4(g)}$ given the following bond energies:$B.E.(N-N) = 159 \ kJ \ mol^{-1}$$B.E.(H-H) = 436 \ kJ \ mol^{-1}$$B.E.(N \equiv N) = 941 \ kJ \ mol^{-1}$$B.E.(N-H) = 398 \ kJ \ mol^{-1}$