Question

Write an equation for the reaction of hydrazine with fluorine gas to produce nitrogen gas and hydrogen fluoride gas. Estimatefor this reaction, using bond energies from Table$\mathbf{\text{13.6}}$.

Short answer

The hydrazide and fluorine gas reaction equation is –

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(l) + 2F}}_{\text{2}}\text{(g)}\to {\text{4HF(g) +N}}_{\text{2}}\text{(g)}$

According to the bond energies the value obtained for$∆H$ , for hydrazine and fluorine gas reaction is, $∆H=-1169kJ$.

Step-by-step solution

Concept Introduction

Bond energy is referred to as average bond enthalpy or simply bond enthalpy, is a measurement of the strength of a chemical bond.

Reaction of hydrazine with fluorine gas

Reaction of hydrazine with fluorine gas to produce nitrogen gas and hydrogen fluoride gas is shown below.

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(l) + 2F}}_{\text{2}}\text{(g)}\to {\text{4HF(g) +N}}_{\text{2}}\text{(g)}$

We can estimate$∆H$from the sum of energies required to break old bonds (positive signs) plus the sum of the energies released in the formation of new bonds (negative signs).

$∆\text{H =}\sum \text{D}$(Bonds broken) $\text{-}\sum \text{D}$(Bonds formed)

Calculation for ∆H

The reaction shown above involves breaking four $\text{N - H}$bonds, two$\text{F - F}$bonds, and one$\text{N - N}$bond. The reaction also involves the formation of$\text{4H - F}$bonds and one triple$\text{N - N}$bond.

${\text{∆H = 4D}}_{\text{N - H}}{\text{+D}}_{\text{N - N}}{\text{+ 2D}}_{\text{F - F}}{\text{- 4D}}_{\text{H - F}}{\text{- 1D}}_{\text{N}\equiv \text{N}}$

$$\begin{gathered} ∆H=4mol\times 391KJ/mol+1mol\times 160KJ/mol \\ +2mol\times 154KJ/mol-4mol\times 565KJ/mol \\ -1mol\times 941KJ/mol \end{gathered}$$

width="118">∆H = - 1169kJ

Therefore, the value of $\text{∆H = - 1169kJ}$.