Question

For the following reactions at constant pressure, predict

$\mathit{i}\mathit{f}\mathbf{}\mathbf{∆}\mathit{H}\mathbf{>}\mathbf{∆}\mathit{E}\mathbf{,}\mathbf{∆}\mathit{H}\mathbf{<}\mathbf{∆}\mathit{E}\mathbf{,}\mathit{o}\mathit{r}\mathbf{∆}\mathit{H}\mathbf{=}\mathbf{∆}\mathit{E}$

1. $\mathbf{2}\mathit{H}\mathit{F}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{F}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$
   
2. ${\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$
   
3. $\mathbf{4}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{5}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{4}\mathit{N}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$
   

Short answer

a).$∆H=∆E$

b).$∆H<∆E$

c).$∆H>∆E$

Step-by-step solution

$$\begin{gathered} ∆H=Enthalpychange, \\ ∆E=Changeininternalenergy. \end{gathered}$$

Step-by-step solution

a)2HF(g)→H2(g)+F2(g)

ΔH = number of gases moles of product - number of gases moles of reactants

Using formula,

$∆H=∆E+∆n_{g}RT$

Where,

$∆n_g=numberofgaseousmolesofproduct-numberofgaseousmolesofreactants$

So, For the given reaction,

$∆n_g=1+1-2=0$

Therefore,$∆H=∆E$

b)N2(g)+3H2(g)→2NH3(g)

For this reaction,

$$\begin{gathered} ∆n_g=2-(1+3)=-2 \\ ∆H=∆E+∆n_{g}RT=∆E-2RT \\ ∆H<∆E \end{gathered}$$

c) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

For this reaction,

$$\begin{gathered} ∆n_g=(4+2)-(4+1)=1 \\ ∆H=∆E+∆n_{g}RT=∆E+RT \\ ∆H>∆E \end{gathered}$$