Question

Consider the equations $:$
$C + {O_2} \to C{O_2};\Delta {H^ \circ } = - xkJ$
$2CO + {O_2} \to 2C{O_2};\Delta {H^ \circ } = - ykJ$
What is the heat of formation of $CO?$
(i) $\dfrac{{y - 2x}}{2}kJ$
(ii) $\left( {2x - y} \right)kJ$
(iii) $\dfrac{{2x - y}}{2}kJ$
(iv) $\left( {y - 2} \right)kJ$

Answer 1

Try to apply Hess’s law. Since we need to find the heat of formation of $CO$ try to express the given equations such that on the product side $CO$ molecule is present. Balance the reactions. Reverse any reaction if required. If any reaction is reversed the magnitude of $\Delta {H^ \circ }$ will remain the same , just its sign will be reversed, also if any reaction is divided by any number the$\Delta {H^ \circ }$of the corresponding reaction will be divided by the same. Add the required reactions for obtaining $CO$ on the product side. Add the $\Delta {H^ \circ }$ of the corresponding reactions and obtain the heat of formation of $CO$ .

Complete step-by-step solution:
Hess’s law states that the change in enthalpy of the overall process is equal to the summation of the enthalpy change of the intermediate steps associated with the process.
Now, the formation of $CO$ can be expressed by the equation
$C + \dfrac{1}{2}{O_2} \to CO.............(1)$
The given equations are $:$
$C + {O_2} \to C{O_2};\Delta {H^ \circ } = - xkJ..........(2)$
$2CO + {O_2} \to 2C{O_2};\Delta {H^ \circ } = - ykJ...........(3)$
Reversing equation $(3)$ we get
$2C{O_2} \to 2CO + {O_2};\Delta {H^ \circ } = ykJ..........(4)$
Since the reaction is reversed the magnitude of $\Delta {H^ \circ }$ will remain the same the sign will be reversed.
Dividing equation $(4)$ by $2$ we get
$C{O_2} \to CO + \dfrac{1}{2}{O_2};\Delta {H^ \circ } = \dfrac{y}{2}kJ..........(5)$
Since the equation is divided by two, $\Delta {H^ \circ }$ is also divided by two.
Now adding equations $(2)$ and $(5)$ we get
$C + \dfrac{1}{2}{O_2} \to CO$ which is the desired equation.
So according to Hess’s law we will now add the $\Delta {H^ \circ }$values of the reactions corresponding to $(2)$ and $(5)$ in order to obtain the heat of formation of $CO$ .
Therefore the heat of formation of $CO$ is given by
$\Delta {H^ \circ }_f(CO) = \left( { - x + \dfrac{y}{2}} \right)kJ = \dfrac{{ - 2x + y}}{2}kJ$

Hence the correct answer is (i) $\dfrac{{y - 2x}}{2}kJ$.

Note: You must take care of the sign of the $\Delta {H^ \circ }$ value provided in the equations. Remember to change the sign when you are reversing the given reaction. Also remember to divide the $\Delta {H^ \circ }$ value of a given reaction by the same number with which you are dividing the equation.