Question

The gas most commonly used in welding is acetylene, \({{\rm{C}}_2}{{\rm{H}}_2}(g)\). When acetylene is burned in oxygen, the reaction is

\({{\rm{C}}_2}{{\rm{H}}_2}(g) + \frac{s}{2}{{\rm{O}}_2}(g) \to 2{\rm{C}}{{\rm{O}}_2}(g) + {{\rm{H}}_2}{\rm{O}}(g)\)

(a) Using data from Appendix D, calculate \(\Delta {H^*}\) for this reaction.

(b) Calculate the total heat capacity of\(2.00\;{\rm{mol}}\),\({\rm{C}}{{\rm{O}}_2}(g)\) and \(1.0D\)mol \({{\rm{H}}_2}{\rm{O}}(g)\), using \({C_p}\left( {{\rm{C}}{{\rm{O}}_2}(g)} \right) = 37{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\) and \({C_{\rm{p}}}\left( {{{\rm{H}}_2}{\rm{O}}(g)} \right) = 36\;{\rm{J}}\;{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\).

(c) When this reaction is performed in an open flame, almost all the heat produced in part (a) goes to increase the temperature of the products. Calculate the maximum flame temperature that is attainable in an open flame burning acetylene in oxygen. The actual flame temperature would be lower than this because heat is lost to the surroundings.

Short answer

(a) The value of \(\Delta H = - 1255.57\;{\rm{kJ}}\).

(b) The total heat capacity is, \({C_P} = 110\;{\rm{J}}\;{{\rm{K}}^{ - 1}}\;{\rm{mol}}\).

(c) The maximum flame temperature is, \({T_{\max }} = 11116.3\;{\rm{K}}\).

Step-by-step solution

Given data

The given reaction is, \({{\rm{C}}_2}{{\rm{H}}_2}(g) + 5/2{{\rm{O}}_2}(g) \to 2{\rm{C}}{{\rm{O}}_2}(g) + {{\rm{H}}_2}{\rm{O}}(g)\).

The enthalpies of formation:

\(\begin{aligned}{}\Delta {H_f}\left( {{{\rm{C}}_2}{{\rm{H}}_2},g} \right) = 226.73\;{\rm{kJ}}\;{\rm{mol}}^{-1}\\\Delta {H_f}\left( {{{\rm{H}}_2}{\rm{O}},g} \right) = - 241.82\;{\rm{kJ}}\;{\rm{mol}}^{- 1}\\\Delta {H_f}\left( {{{\rm{O}}_2},g} \right) = 0\;{\rm{kJ}}\;{\rm{mol}}^{- 1}\end{aligned}\)

The heat capacities are shown below.

\(\begin{aligned}{}{C_p}\left( {{{\rm{H}}_2}{\rm{O}},g} \right) = 36\;{{\rm{J}}^{ - 1}}\;{\rm{mol}}\\{C_P}\left( {{\rm{C}}{{\rm{O}}_2},g} \right) = 37{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mol}}\end{aligned}\)

The numbers of moles of each molecule are given below.

\(\begin{aligned}{}n\left( {{{\rm{H}}_2}{\rm{O}}} \right) = 1.00\;{\rm{mol}}\\n\left( {{\rm{C}}{{\rm{O}}_2}} \right) = 2.00\;{\rm{mol}}\end{aligned}\)

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\(\Delta H\)and is termed as the reaction enthalpy.

The reaction enthalpy is calculated as:

Mathematically,\(\Delta H = \sum \Delta {H_f}({\rm{ products }}) - \sum \Delta {H_f}({\rm{ reactants }})\).

Where, \(\Delta H\)is change in enthalpy/

Calculate change in enthalpy

(a) The enthalpy change formula is:

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

Put the value of given date in above equation.

\(\begin{aligned}{}\Delta H &= 2 \times \Delta {H_f}\left( {{\rm{C}}{{\rm{O}}_2},g} \right) + \Delta {H_f}\left( {{{\rm{H}}_2}{\rm{O}},g} \right) - \left( {\Delta {H_f}\left( {{{\rm{C}}_2}{{\rm{H}}_2},g} \right) + \frac{5}{2}\Delta {H_f}\left( {{{\rm{O}}_2},g} \right)} \right)\\\Delta H &= 2 \times ( - 393.51) + ( - 241.82) - (226.73 + 0)\\\Delta H &= - 1255.57\;{\rm{kJ}}\end{aligned}\)

Calculate the total heat capacity of the reaction

(b) The total heat capacity in this case will be calculated by the formula.

\({C_p} = n\left( {{\rm{C}}{{\rm{O}}_2}} \right) \times {C_p}\left( {{\rm{C}}{{\rm{O}}_2}} \right) + n\left( {{{\rm{H}}_2}{\rm{O}}} \right) \times {C_p}\left( {{{\rm{H}}_2}{\rm{O}}} \right)\)

Put the value of given date in above equation.

\({C_p} = 110\;{\rm{J}}\;{{\rm{K}}^{ - 1}}\;{\rm{mol}}\)

Calculate maximum temperature

(c) The heat is calculated by, \(q = {C_p}\Delta T\).

So, the temperature change would be, \(\Delta T = \frac{q}{{{C_p}}}\).

Where \(q\) is the total heat and\({C_p}\) is the total heat capacity which is calculated in step 3 and 4.

So, change in temperature becomes:

\(\begin{aligned}{}\Delta T = \frac{{1255.57 \times {{10}^3}}}{{110}}\\\Delta T = 11414.3\;{\rm{K}}\end{aligned}\)

For calculation of maximum temperature, subtract the initial temperature is 25⁰C or \(298\;{\rm{K}}\)from change in temperature.

\(\begin{aligned}{}{T_{\max }} = 11414.3 - 298\\{T_{\max }} = 11116.3\;{\rm{K}}.\end{aligned}\)