Question

The heats of formation of $C{{O}_{2}}(g)$and ${{H}_{2}}O(l)$ are -394 kJ/mole and -285.8 kJ/mole respectively. Using the data for the following combustion reaction, calculate the heat of formation of ${{C}_{2}}{{H}_{2}}(g)$.
\[2{{C}_{2}}{{H}_{2}}(g)+5{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+2{{H}_{2}}O(l),\,\Delta {{H}^{0}}=-2601\,kJ\]
A. -238.6
B. 253.2
C. 238.7
D. 226.7

Answer 1

Hint: Enthalpy or heats of formation is defined as the energy change that happens, when 1 mole of a compound is formed from the constituent atoms of that compound. It takes place in a standard condition of temperature of $25{}^\circ C$and a pressure of 1 bar.

Complete answer:
We have been given the heats of formation of carbon dioxide, which is, -394 kJ/mole as,
\[C+{{O}_{2}}(g)\to C{{O}_{2}}(g),\,\Delta {{H}_{1}}=-394kJ/mole\,\,...(1)\]
And that of water molecule as, -285.8 kJ/mol, from the reaction,
\[{{H}_{2}}+\dfrac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l),\,\Delta {{H}_{2}}=-285.8kJ/mole\,\,\,...(2)\]
Now, we are also given the enthalpy for the reaction,
\[2{{C}_{2}}{{H}_{2}}(g)+5{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+2{{H}_{2}}O(l),\,\Delta {{H}^{0}}=-2601\,kJ\,\,\,...(3)\]
We have to find out the heat of formation of ethylene gas which is, ${{C}_{2}}{{H}_{2}}(g)$. Its reaction is,
\[2C+{{H}_{2}}\to {{C}_{2}}{{H}_{2}}(g)\]
Now, adding the equations, (1) and (2) and balancing them by multiplying equation (1) $\times 2$ we will get the enthalpy of formation of water and carbon dioxide, as,
\[2C+{{O}_{2}}(g)+{{H}_{2}}+\dfrac{1}{2}{{O}_{2}}(g)\to 2C{{O}_{2}}+{{H}_{2}}O,\,\Delta {{H}_{3}}=-1073.8kJ\,\,\,...(4)\]
Now, to obtain ethylene, and to cancel out the products we will reverse the equation number (3) and label it as equation (5), as,
\[4C{{O}_{2}}(g)+2{{H}_{2}}O(l)\to 2{{C}_{2}}{{H}_{2}}(g)+5{{O}_{2}}(g),\,\Delta {{H}^{0}}=2601\,kJ\,\,\,...(5)\]
Now, to obtain ethylene gas, we will first divide equation (5) by 2, and obtain equation (6),
\[2C{{O}_{2}}(g)+{{H}_{2}}O(l)\to {{C}_{2}}{{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g),\,\Delta {{H}^{0}}=1300.5\,kJ\,\,\,...(6)\]
Now, to obtain the equation of formation of ethylene we will add equation (4) and (6), and cancel out the common reactants and products to obtain, the equation of forming ethylene, and thus the heat of formation of ethylene will be calculated as,
\[2C+{{H}_{2}}\to {{C}_{2}}{{H}_{2}}(g),\,\,\Delta {{H}^{0}}_{f}=226.7\,kJ\]
Hence, the heat of formation of ${{C}_{2}}{{H}_{2}}(g)$gas is 226.7 kJ.

So, option D is correct.

Note:
When operations are done between equations, then addition, or subtraction, or multiplication and division, is done for equations as well as their enthalpies of formations and reactions, this simple operation, gives us the desired values.