Chapter 6, Problem 134IP

Nitromethane, CH₃NO₂, can be used as a fuel. When the liquid is burned, the (unbalanced) reaction is mainly

CH₃NO₂(l) + O₂(g) → CO₂(g) + N₂(g) + H₂O(g)

1. a. The standard enthalpy change of reaction (ΔH°_van ) for the balanced reaction (with lowest whole-number coefficients) is −1288.5 kJ. Calculate ΔH_f⁰ for nitromethane.
2. b. A 15.0-L flask containing a sample of nitromethane is filled with O₂ and the flask is heated to 100.°C. At this temperature, and after the reaction is complete, the total pressure of all the gases inside the flask is 950. torr. If the mole fraction of nitrogen (χ_nitrogen) is 0.134 after the reaction is complete, what mass of nitrogen was produced?

Interpretation Introduction

Interpretation:

Considering the given equation the enthalpy of formation for nitromethane and the mass of nitrogen produced when given volume of sample is stored under $\text{950 torr}$ at ${\text{100}}^{\circ }\text{C}$ should be determined.

Concept Introduction:

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. The standard enthalpy of the reaction is equal to the subtraction of total enthalpy for formation product with the total enthalpy for formation of the reactants which is given as ${\text{ΔH}}^{\circ }{}_{\text{rxn}}=\sum n{\text{ΔH}}^{\circ }{}_f\left(\text{products}\right)-\sum m{\text{ΔH}}^{\circ }{}_f\left(\text{reactants}\right)$.

Ideal Gas equation: It is the equation which equals the pressure and volume product of one mole of gas to the temperature and gas constant which is $\text{PV = nRT}$.

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Limiting Reagent: A reactant is considered as limiting reagent when it limits the product formation that depend on the amount of that reactant.

Answer to Problem 134IP

The standard enthalpy of formation ${\text{ΔH}}_{\text{f}}$ for the given product Nitro-methane is $\text{- 434}\text{.4kJ/mol}$.

Explanation of Solution

To determine: The ${\text{ΔH}}_{\text{f}}$ value for given product and the mass of nitrogen produced under the given conditions.

Given,

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+O}}_{\text{2}}{}_{\left(\text{g}\right)}\to {\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}+{\text{N}}_{\text{2}\left(\text{g}\right)}+{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ \Delta {\text{H}}^{\circ }{}_{\text{rxn}}=-1288.5\text{kJ}\end{array}$

Analyse the given equation and balance it.

The unbalanced equation is ${\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+O}}_{\text{2}}{}_{\left(\text{g}\right)}\to {\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}+{\text{N}}_{\text{2}\left(\text{g}\right)}+{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$.

Therefore, the balanced equation where the number of moles of product and the reactant on both side of the equation should be equal is depicted as follows ${\text{4CH}}_{\text{3}}{\text{NO}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+3O}}_{\text{2}}{}_{\left(\text{g}\right)}\to 4{\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}+2{\text{N}}_{\text{2}\left(\text{g}\right)}+6{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$.

Now, the enthalpy of formation for the given product is calculated as follows,

$\begin{array}{l}{\text{ΔH}}^{\text{o}}{}_{\text{rxn}}\text{=}{\text{ånΔH}}^{\text{o}}{}_{\text{f}}\left(\text{products}\right){\text{-åmΔH}}^{\text{o}}{}_{\text{f}}\left(\text{reactants}\right)\\ \text{-1288}\text{.5kJ}\text{=}\text{4}\left(\text{-393}\text{.5kJ}\right)\text{+2}\left(\text{0}\right)\text{+6}\left(\text{-242kJ}\right)\text{- 4}\left({\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}\right)-3\left(0\right)\\ 4\left({\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}\right)=-1574\text{kJ + 0 -1452 - 0+1288}\text{.5kJ}\\ \left({\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}\right)=\frac{-\text{1737}\text{.5}}{4}\\ =-434.4\text{kJ/mol}\end{array}$

Conclusion

The enthalpy of formation for the given product nitromethane was determined.

Interpretation Introduction

Interpretation:

Considering the given equation the enthalpy of formation for nitromethane and the mass of nitrogen produced when given volume of sample is stored under $\text{950 torr}$ at ${\text{100}}^{\circ }\text{C}$ should be determined.

Concept Introduction:

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. The standard enthalpy of the reaction is equal to the subtraction of total enthalpy for formation product with the total enthalpy for formation of the reactants which is given as ${\text{ΔH}}^{\circ }{}_{\text{rxn}}=\sum n{\text{ΔH}}^{\circ }{}_f\left(\text{products}\right)-\sum m{\text{ΔH}}^{\circ }{}_f\left(\text{reactants}\right)$.

Ideal Gas equation: It is the equation which equals the pressure and volume product of one mole of gas to the temperature and gas constant which is $\text{PV = nRT}$.

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Limiting Reagent: A reactant is considered as limiting reagent when it limits the product formation that depend on the amount of that reactant.

Answer to Problem 134IP

The mass of nitrogen produced from the given reaction under given conditions is $\text{12}\text{.5g}$.

Explanation of Solution

Given,

$\text{15L}$ of flask contains nitromethane at ${\text{100}}^{\text{o}}\text{C}$.

The total pressure of gas is $\text{950 torr }$

Given mole fraction of nitrogen is $\text{0}\text{.134}$.

Analysing the given reaction shows us that 4 moles of ${\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}{}_{\left(\text{l}\right)}$ results in 12 moles of gas products therefore 1 mole of ${\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}{}_{\left(\text{l}\right)}$ gives 4 moles of gas products.

$\begin{array}{l}\text{PV = nRT}\\ \text{n =}\frac{\text{PV}}{\text{RT}}\\ =\frac{\text{1}\text{.25atm}\times \text{15L}}{\text{0}\text{.08206}\times \text{373}}\\ =0.612\text{mol}\\ \text{where,}\\ \text{P = 950 torr}\frac{1\text{atm}}{760\text{torr}}=1.25\text{atm}\\ {\text{T = 100}}^{\text{o}}\text{C}\text{+}\text{273K}=\text{373K}\end{array}$

$\begin{array}{l}\text{=0}\text{.612 mol gas}\left(\frac{\text{1}\text{mol}\text{CH3NO2}}{\text{3}\text{mol}\text{gas}}\right)\\ \text{=0}\text{.204}\text{mol}\\ =\left(0.204\text{mol}\right)\left(61.0400\text{6g/mol}\right)\\ =12.5\text{g}\\ {\text{where, molecular mass of CH}}_{\text{3}}{\text{NO}}_{\text{2}}\text{ = 61}\text{.04006g/mol}\text{.}\end{array}$

Therefore, the mass of nitrogen produced during the given reaction under given conditions is $12.5\text{g}$.

Conclusion

The mass of nitrogen produced during the given reaction was determined.