Question

Enthalpy of combustions for the following reactions are given.
$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right)$ ; $\Delta {H_c} = - 394KJmo{l^{ - 1}}$
$${H_2}\left( g \right) + \dfrac{1}{2}{O_2} \to {H_2}O$$ ; $$\Delta {H_c} = - 286 \,KJmo{l^{ - 1}}$$
$CH_3OH(l)+\dfrac{3}{2}O_2(g)\to CO_2(g)+2H_2O(l)$ ; $$\Delta {H_c} = - 726 \,KJmo{l^{ - 1}}$$
Which of the following expressions gives the standard enthalpy change of formation of methanol?
A. $ - 394 + \left( { - 286} \right) - \left( { - 726} \right)$
B. $ - 394 + ( - 286 \times 2) - 726$
C. $ - 394 + ( - 286 \times 2) - ( - 726)$
D. $ - 726 - ( - 394) - ( - 286 \times 2)$

Answer 1

Hint: Hess’ law used in thermodynamics states that the change of enthalpy in a chemical reaction is independent of the path taken between initial and final states. If the enthalpy change for each equation is known, the resultant enthalpy change is the net reaction.

Complete step by step solution:
Hess’ law used in thermodynamics states that the change of enthalpy in a chemical reaction is independent of the path taken between initial and final states. If the enthalpy change for each equation is known, the resultant enthalpy change is the net reaction. This is because Enthalpy is a state function that is, it is not dependent on the path taken. Enthalpy of a system is the total of internal energy$\left( U \right)$ and product of pressure and volume. Change in enthalpy $\left( {\Delta H} \right)$of the system tells us the amount of heat absorbed or evolved by the system.
The heat of combustion is the amount of heat released when $1$ mole of a compound is completely burnt. Since it is an exothermic reaction $\Delta H$ is negative. According to the given question
The heat of combustion of carbon is $\Delta {H_c} = - 394\,KJmo{l^{ - 1}}$
$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right)$………$\left( 1 \right)$
The heat of combustion of hydrogen is $$\Delta {H_c} = - 286\,KJmo{l^{ - 1}}$$
$${H_2}\left( g \right) + \dfrac{1}{2}{O_2} \to {H_2}O$$
Multiplying equation $\left( 2 \right)$ by $2$
$$2 \times \left[ {{H_2}\left( g \right) + \dfrac{1}{2}{O_2} \to {H_2}O} \right]$$……$\left( 2 \right)$
The heat of combustion of hydrogen also gets double that Is $\Delta {H_c} = - 576\,KJmo{l^{ - 1}}$
The heat of combustion of methanol is $\Delta {H_c} = - 726\,KJmo{l^{ - 1}}$
$CH_3OH(l)+\dfrac{3}{2}O_2(g)\to CO_2(g)+2H_2O(l)$
Reversing the above equation
$CO_2(g)+2H_2O(l) \to CH_3OH(l)+\dfrac{3}{2}O_2(g)$ …………..$\left( 3 \right)$
Since the equation is reversed, the sign of enthalpy change also gets reversed $\Delta {H_c} = 726\,KJmo{l^{ - 1}}$
Adding equations $\left( 1 \right)$ , $\left( 2 \right)$ and $\left( 3 \right)$, to find the enthalpy change of formation of methanol
$$C(s) + 2{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to C{H_3}OH\left( l \right)$$
$ \Rightarrow \Delta {H_{f = }} - 394 + ( - 286 \times 2) - ( - 726)$

Therefore, the correct option is (C)

Note:
Hess law of constant heat summation is a direct consequence of the enthalpy being a state function. It is a very useful equation used to calculate the change in enthalpy from a large range of data. As most of the substances cannot be prepared directly from the elements, heat of formation is seldom determined by direct measurement.