Question

Calculate the standard enthalpy change $∆H^{\circ }$at 25⁰Cfor the reaction ${\mathbf{\text{N}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{\text{O}}}_{\mathbf{1}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{\text{NO}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}$ using the standard enthalpies of formation (AH_f⁰) ofreactants and products at 25⁰Cfrom Appendix D.

Short answer

The standard enthalpy change $\Delta H$is $-555.93\text{kJ}$.

Step-by-step solution

Given data

The enthalpy changes to make diamond from graphite is $1.88{\text{kJmol}}^{-1}$.

${\text{N}}_2{\text{H}}_2\left(\ell \right)+3{\text{O}}_1\left(g\right)\to 2{\text{NO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(\ell \right)$

$∆H_f^{\circ }$ Value of ${\text{NO}}_2=33.18{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$Value of ${\text{H}}_2\text{O}=-285.83{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$Value of ${\text{N}}_2{\text{H}}_4=50.63{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$Value of ${\text{O}}_2=0$.

$∆H_f^{\circ }$Represents the standard enthalpy of molecules

 Step 2: Concept of Enthalpy

Step 3: Calculate the value of

Negative enthalpy change represents the exothermic reaction when energy is released from the reaction and positive enthalpy change represents an endothermic reaction in which energy is taken in from surroundings.

Calculate the value of∆H

The given data is:

$∆H_f^{\circ }$Value of ${\text{NO}}_2=33.18{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$Value of $H_{2}O=-285.83kJmo{l}^{-1}$.

$∆H_f^{\circ }$Value of ${\text{N}}_2{\text{H}}_4=50.63{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$Value of ${\text{O}}_2=0$.

$∆H_f^{\circ }$Represents the standard enthalpy of molecules.

The given expression is ${\text{N}}_2{\text{H}}_2\left(\ell \right)+3{\text{O}}_1\left(g\right)\to 2{\text{NO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(\ell \right)$.

The enthalpy changes of reaction ($∆H$) is as $\Delta H=n\Delta H_f^e\left(Products\right)-n\Delta H_f^o\left(Reactants\right).$.

$$\begin{gathered} ∆H=2∆H_f^{\circ }N{O}_2\left(g\right)+2∆H_f^{\circ }{H}_{2}O\left(l\right)-∆H_f^{\circ }{N}_2{H}_4\left(l\right)+3∆H_f^{\circ }{O}_2\left(g\right) \\ =2mol\times 33.18kJmo{l}^{-1}+2mol\times -285.83kJmol{m}^{-1}-1mol\times 50.63kJmo{l}^{-1}+3mol\times 0 \\ =66.36kJ-571.66kJ-50.63kJ \\ ∆H=-555.93kJ \end{gathered}$$