Question

Given the following standard heats of reactions:
(a)heat of formation of water= -68.3kcal, (b) heat of combustion of ${{\text{C}} _ {2}} {{\text{H}} _ {2}} $= -310.6 kcal and (c) heat of combustion of ethylene= -337.2 kcal. Calculate the heat of the reaction for the hydrogenation of acetylene at constant volume and at 25$^ {\circ} \text{C}$.

Answer 1

Hint: First find out the heat of reaction (may be defined as the change in the enthalpy i.e. the total heat content of the system) of acetylene from the data of the given reactions in the statement by adding or subtracting those reactions and then by using ideal gas relation of enthalpy and heat of energy, you easily calculate $\Delta \text{E}$ at constant volume.

Complete step by step answer:
By the heat of reaction, we simply mean the enthalpy change when 1 mol of any substance reacts with excess of the hydrogen atom under conditions of atmospheric pressure and room temperature. Now, from above given data, the enthalpy of the reaction comes out to be:
heat of formation of water = -68.3kcal (given), i.e. The reaction is:
${{\text{H}} _ {2}} \text {+} \dfrac {1}{2} {{\text{O}} _ {2}} \to \text {} {{\text{H}} _ {2}} \text{O}$ ----------(i)
$\Delta \text{H}$=−68.3Kcal
Heat of combustion of ${{\text{C}} _ {2}} {{\text{H}} _ {2}} $= - 310.6 kcal (given), i.e. The reaction is:

${{\text{C}}_{2}}{{\text{H}}_{2}}\text{+}\dfrac{5}{2}{{\text{O}}_{2}}\to \text{ 2C}{{\text{O}}_{2}}\text{+}{{\text{H}}_{2}}\text{O}$ ------------(ii)

$\Delta \text{H}$=−310.6Kcal
Heat of combustion of ethylene = -337.2 kcal, i.e. the reaction is:
 ${{\text{C}}_{2}}{{\text{H}}_{4}}\text{+3}{{\text{O}}_{2}}\to \text{ 2C}{{\text{O}}_{2}}\text{+2}{{\text{H}}_{2}}\text{O}$ -----------(iii)
 $\Delta \text{H}$=−337.2Kcal

Heat of hydrogenation of acetylene has to be found i.e. $\Delta \text{H}$has to be calculated and its reaction is:
${{\text{C}} _ {2}} {{\text{H}} _ {2}} \text {+} {{\text{H}} _ {2}} \to {{\text{C}} _ {2}} {{\text{H}} _ {4}} $ $\Delta \text{H}$=? ---------(iv)

Arrange the above equation i.e.(i), (ii)and (iii) in such a way that we get the equation same as equation(iv). So, we observe that on adding the equations (i) and (ii) and subtracting equation (iii), we get the same equation as (iv), i.e.
${{\text{C}}_{2}}{{\text{H}}_{2}}\text{+}{{\text{H}}_{2}}\to {{\text{C}}_{2}}{{\text{H}}_{4}}$ $\Delta \text{H}$= −41.7Kcal​

Thus, this is the heat of reaction at constant pressure. So, thus the heat of the reaction at constant volume can be calculated by using the equation of enthalpy and heat of reaction relation of ideal gas equation. So, we get:
$\Delta \text{H}$=$\Delta \text{E}$+$\Delta {{\text{n}} _{g}} $RT ----------(v)

Where $\Delta \text{H}$is the earthly, $\Delta \text{E}$is the heat of the reaction, $\Delta {{\text{n}} _{g}} $ is the no of moles gaseous reactants minus the no of moles of the gaseous products i.e.
$\Delta {{\text{n}} _{g}} $= moles of gaseous products - moles of gaseous reactants
R is the gas constant and whose value in kcal/ mole is 0.002. and T is the temperature.
$\Delta \text{H}$= -41.7 k call
R= 0.002
T =273+25
 = 298K

For $\Delta {{\text{n}} _{g}} $, consider the following reaction, we get:
 ${{\text{C}} _ {2}} {{\text{H}} _ {2}} \text {+} {{\text{H}} _ {2}} \to {{\text{C}} _ {2}} {{\text{H}} _ {4}} $
$\Delta {{\text{n}} _{g}} $=1−2
 =−1

So, putting all these values in equation (v), we get:
 $\Delta \text{E}$=−41.7−(−1×0.002×298)
 $\Delta \text{E}$=−41.104 Kcal
Therefore, the heat of the reaction for the hydrogenation of acetylene at constant volume and at 25$^ {\circ} \text{C}$ is -41.104 kcal.
So, the correct answer is “Option C”.

Note: The enthalpy of the reaction $\Delta \text{H}$ always depends on the temperature and is independent of the pressure. And $\Delta \text{H}$ is always negative for exothermic reactions i.e.in which the heat is released and on the other $\Delta \text{H}$ is negative for those reactions which absorb from surrounding that for endothermic reactions.