Question
The standard enthalpy changes for the combustion of carbon and carbon monoxide are shown below.
\[\begin{array}{*{20}{l}} {{\text{C(s)}} + {{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}}}&{\Delta H_{\text{c}}^\Theta = – 394{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}} \\ {{\text{CO(g)}} + \frac{1}{2}{{\text{O}}_2}{\text{(g)}} \to {\text{C}}{{\text{O}}_2}{\text{(g)}}}&{\Delta H_{\text{c}}^\Theta = – 283{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}} \end{array}\]
What is the standard enthalpy change, in kJ, for the following reaction?
\[{\text{C(s)}} + \frac{1}{2}{{\text{O}}_2}{\text{(g)}} \to {\text{CO(g)}}\]
A. –677
B. –111
C. +111
D. +677
▶️Answer/Explanation
B
As per Hess’s law :
If reactants react to form products not in a single step but in a number of consecutive steps involving many intermediary products, the sum of all the reactants, products and the corresponding energy changes will give the reactant, products and heat energy changes of the overall reaction. So, like molecules, heat energy changes also can be subjected to mathematical operations.
C(s)+O2(g)→CO2(g) (Equation 1); ΔH1= -394 kJ mol-1
CO(g)+\(\frac{1}{2}\)O2(g)→CO2(g) (Equation 2); ΔH2= -283 kJ mol-1
Here, if we subtract the second equation from the first equation, i.e. (Equation 1 – Equation 2 ), we get,
\[{\text{C(s)}} + \frac{1}{2}{{\text{O}}_2}{\text{(g)}} \to {\text{CO(g)}}\] (Equation 3); ΔH3
which is the desired equation.
Now, as per Hess’s law, standard enthalpy change, in kJ, for the Equation 3,
ΔH3= ΔH(Equation 1) – ΔH(Equation 2 )
ΔH3=ΔH2 – ΔH1
ΔH3= -394 – (-283) = -111 kJ
Question
B. −788−286−1301
C. +788+286−1301
D. +788+286+1301
▶️Answer/Explanation
A
Hess law states that :
If the reactants and products of a required chemical reaction can be obtained by the summation of many other chemical reactions, the enthalpy of the required reaction of reactants to the products also can be obtained by the sum of the enthalpy changes of all those chemical reactions.
Here, ethyne can be formed from direct equation Y.
Also, it can be formed by following sequence of equations as represented: Arrow having -788 kJ energy followed by -286 kJ followed by reverse of arrow representing -1301 kJ.
Y = (-788) + (-286) – (-1301)
Y = -788-286+1301
Hence, enthalpy of the reaction :
2C(graphite) + H2(g) + \(\frac{5}{2}\)O2(g) → C2H2(g) + \(\frac{5}{2}\)O2(g) is -788-286+1301.
ΔH of a reaction is calculated by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products,
Here, ΔH of C(graphite) is zero and also of H2(g) is zero since all elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.) have a standard enthalpy of formation of zero, as there is no change involved in their formation.
So, [ΔH (C2H2(g)) + ΔH (\(\frac{5}{2}\)O2(g)) ] – [ 0 + 0 + ΔH (\(\frac{5}{2}\)O2(g))] = ΔH (Y)
ΔH (C2H2(g)) = -788-286+1301