Question

Nitro methane, \({\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{O}}_2}\), is a good fuel. It is a liquid at ordinary temperatures. When the liquid is burned, the reaction involved is chiefly

\(2{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{O}}_2}(\ell ) + \frac{3}{2}{{\rm{O}}_2}(g) \to \mathop 2\limits_2 {\rm{C}}{{\rm{O}}_2}(g) + {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}{\rm{O}}(g)\). The standard enthalpy of formation of liquid nitro methane at 25⁰C is \( - 112\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\), other relevant values can be found in Appendix D.

(a) Calculate the enthalpy change in the burning of \(1\;{\rm{mol}}\) liquid nitro methane to form gaseous products at 25⁰C. State explicitly whether the reaction is endothermic or exothermic.

(b) Would more or less heat be evolved if gaseous nitro methane were burned under the same conditions? Indicate what additional information (if any) you would need to calculate the exact amount of heat, and show just how you would use this information.

Short answer

(a) The enthalpy changes for the reaction\(2{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{O}}_2}(\ell ) + \frac{3}{2}{{\rm{O}}_2}(g) \to \mathop 2\limits_2 {\rm{C}}{{\rm{O}}_2}(g) + {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}{\rm{O}}(g)\)is,\( - 1400.5\;{\rm{J}}\).

(b)The standard enthalpy of formation of gaseous nitro methane is required.

Step-by-step solution

Given data

The given reaction is,\(2{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{O}}_2}(\ell ) + \frac{3}{2}{{\rm{O}}_2}(g) \to \mathop 2\limits_2 {\rm{C}}{{\rm{O}}_2}(g) + {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}{\rm{O}}(g)\).

The enthalpy changes for above reaction from the element at 25⁰C is \( - 112\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\).

Concept of change in enthalpy of reaction

Enthalpy change is the standard enthalpy of formation, which has been determined for a vast number of substances. In any general chemical reaction, the reactants undergo chemical changes and combine to give products.

It can be represented by the following equation:

\({\mathop{\rm Re}\nolimits} ac\tan t \to \Pr oduct\)

For any such reaction, the change in enthalpy is represented as∆Hand is termed as the reaction enthalpy. The reaction enthalpy is calculated as:

Mathematically,\[\Delta H = \Delta {H_1} + \Delta {H_2}\].

Where,∆H₁ is change in enthalpy of reactant and ∆H₂ is change in enthalpy of product.

Calculation of enthalpy change 

(a) The standard enthalpies of formation:

\(\begin{array}{c}\Delta {H_f}\left( {{\rm{C}}{{\rm{O}}_2},g} \right) = - 393.51\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\\\Delta {H_f}\left( {{{\rm{H}}_2}{\rm{O}},g} \right) = - 241.82{\rm{kJmo}}{{\rm{l}}^{ - 1}}\\\Delta {H_f}\left( {{{\rm{O}}_2},g} \right) = 0\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\\\Delta {H_f}\left( {\;{{\rm{N}}_2},g} \right) = 0\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\end{array}\)

The overall enthalpy change for a reaction is written as:

\(\Delta H = \sum \Delta H({\rm{ products }}) - \sum \Delta H({\rm{ reactants }})\)

In our case the equation would look like this:

\(\begin{array}{c}\Delta H = 2\Delta {H_f}\left( {{\rm{C}}{{\rm{O}}_2},g} \right) + 3\Delta {H_f}\left( {{{\rm{H}}_2}{\rm{O}},g} \right) - \Delta {H_f}\left( {{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{O}}_2},l} \right)\\\Delta H = 2 \times ( - 393.51) + 3 \times ( - 241.82) - ( - 112)\\\Delta H = - 1400.5\;{\rm{J}}\end{array}\)

Determination of reaction type

Determining if the reaction is exothermic or endothermic

Our enthalpy change value is negative.

This means that the heat is being released from the system to the surroundings, making the reaction exothermic.

Determine more or less heat be evolved if gaseous nitro methane were burned under the same conditions

(b) Gaseous substances burn more easily than liquid substances.

This means less energy is needed to make them burn.

This would also mean that when they burn, they burn quicker and stronger.

Therefore, the burning of gaseous nitro methane will release more heat.

Determine additional information (if any) that need to calculate the exact amount of heat

To calculate the enthalpy of this process, follow the same steps as used in step 3.

The reaction,\(2{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{O}}_2}(g) + 3/2{{\rm{O}}_2}(g) \to 2{\rm{C}}{{\rm{O}}_2}(g) + {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}{\rm{O}}(g)\).

The enthalpy change equation,\(\Delta H = 2\Delta {H_f}\left( {{\rm{C}}{{\rm{O}}_2},g} \right) + 3\Delta {H_f}\left( {{{\rm{H}}_2}{\rm{O}},g} \right) - \Delta {H_f}\left( {{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{O}}_2},g} \right)\).

The standard enthalpy of formation of gaseous nitro methane is required.