Question

Calculate the heat of hydrogenation of ${C_2}{H_2}$ to ${C_2}{H_4}$.
${H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to {H_2}{O_{(l)}};\Delta H = - 68.32kcal$
${C_2}{H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to 2C{O_{2(g)}} + {H_2}{O_{(l)}};\Delta H = - 310.61kcal$
${C_2}{H_{4(g)}} + 3{O_{2(g)}} \to 2C{O_{2(g)}} + {H_2}{O_{(l)}};\Delta H = - 337.32kcal$
A) $ - 141.61kcal$
B) $ - 41.61kcal$
C) $41.61kcal$
D) None of these

Answer 1

Hint: Hint- The heat of hydrogenation is the heat required to add one hydrogen molecule to the reactant molecule that is in this question is ${C_2}{H_2}$ a molecule. One can analyze the given equations and find out the hydrogenation chemical reaction and the heat values of each reaction.

Complete step by step answer:
1) First of all we will learn about the heat of hydrogenation reaction, the heat of hydrogenation reaction of alkenes is a measurement of the stability present between the carbon-carbon double bonds. As the smaller, the value of heat of hydrogenation reaction of an alkene gets the more stable the double bond is present there.
2) So the general representation for the reaction will be as follows,
\[{C_2}{H_2} + {H_2} \to {C_2}{H_4}\]
3) Now let's analyze the given chemical equations and find the reaction of the heat of hydrogenation,
The first reaction is the formation of reactive oxygen which is,
${H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to {H_2}{O_{(l)}};\Delta H = - 68.32kcal$
In the second reaction, the oxygen is reacted with the ethylene molecule to get the byproducts water and carbon dioxide.
${C_2}{H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to 2C{O_{2(g)}} + {H_2}{O_{(l)}};\Delta H = - 310.61kcal$
In the third reaction, the ethene molecule is reacted with three molecules of the oxygen to give water and carbon dioxide molecules.
${C_2}{H_{4(g)}} + 3{O_{2(g)}} \to 2C{O_{2(g)}} + {H_2}{O_{(l)}};\Delta H = - 337.32kcal$
4) Now we need to get the general reaction for the heat of hydrogenation of ethylene molecules by using these reactions, and we can obtain this by keeping the reaction first and second as it is and reversing the reaction third. By doing this after the cancellation of the equal molecules present on both sides we get the general reaction which is, \[{C_2}{H_2} + {H_2} \to {C_2}{H_4}\]
Let's put this in mathematical form,
$( - 68.32) + ( - 310.61) + ( + 337.32)$ (we have taken the last value as we can get the desired general reaction by reversing the third reaction which also reverses the heat value to positive)
$ = ( - 378.93) + 337.32 = - 41.61kcal$
5) Therefore, the heat of hydrogenation of ${C_2}{H_2}$ to ${C_2}{H_4}$ is $ - 41.61kcal$ .

So, the correct answer is Option C.

Note:
The reactions given in the question always lead to a general reaction of which heat is to be calculated. So, one can always obtain this by equating all the given reactions by reversing the reaction equation where needed and canceling the same molecules on both sides which at the end gives the desired general reaction. One should remember to do calculations with $ + \& - $ the sign of heat values to obtain accurate values.