Question

Calculate \[\Delta H_f^0\] of \[{C_6}{H_{12}}{O_6}\](s) from the following data:
\[\Delta {H_{Comb}}\] of \[{C_6}{H_{12}}{O_6}\](s) = -2816 KJ \[mo{l^{ - 1}}\], \[\Delta H_f^0\] of \[C{O_2}\](g) = -393.5 KJ \[mo{l^{ - 1}}\] and \[\Delta H_f^0\] $H_2O$= -285.9KJ\[mo{l^{ - 1}}\].
A. 1260 KJ\[mo{l^{ - 1}}\]
B. -1260 KJ\[mo{l^{ - 1}}\]
C. -2260 KJ\[mo{l^{ - 1}}\]
D. -3260 KJ \[mo{l^{ - 1}}\]

Answer 1

Hint:To calculate the heat of formation Hess’s law of heat summation is applied where the heat evolved or released during the chemical reaction is calculated by summing the known value for heat of formation or combustion for each step.

Complete step by step answer:
Given,
\[\Delta {H_{Comb}}\] of \[{C_6}{H_{12}}{O_6}\](s) = -2816 KJ \[mo{l^{ - 1}}\]
\[\Delta H_f^0\] of \[C{O_2}\](g) = -393.5 KJ \[mo{l^{ - 1}}\]
\[\Delta H_f^0\] $H_2O$= -285.9KJ\[mo{l^{ - 1}}\].
The reaction for the combustion of one mole glucose is shown below.
\[{C_6}{H_{12}}{O_6}(s) + 6{O_2}(g) \to 6C{O_2}(g) + 6{H_2}O(l)\]
In this reaction, one mole of glucose reacts with six mole of oxygen to form 6 mole of carbon dioxide and six mole of water.
The heat of formation is also known as standard heat of formation, enthalpy of formation and standard enthalpy of formation.
The enthalpy of formation is defined as the amount of heat released or absorbed in the formation of one mole of compound from its constituent elements where the elements are present in their normal standard state.
By using Hess’s law of heat summation, the heat evolved or released during the chemical reaction is calculated by summing the known value for heat of formation or combustion for each step.
The enthalpy of formation for glucose is calculated as shown below.
\[\Delta {H_f}\;of\;{C_6}{H_{12}}{O_6}\]=\[6 \times \Delta {H_f} of\;C{O_2} + 6 \times \Delta {H_f} of\;{H_2}O - \Delta {H_{combus}} of\;{C_6}{H_{12}}{O_6}\]
To calculate the enthalpy of formation for glucose, substitute the values in above expression as shown below.
\[\Delta {H_f}\;of\;{C_6}{H_{12}}{O_6}\]=\[6 \times ( - 393.5) + 6 \times ( - 285.9) - ( - 2816)\]
\[\Delta {H_f}\;of\;{C_6}{H_{12}}{O_6}\]=-1260 KJ \[mo{l^{ - 1}}\]
Thus, the heat of formation of glucose is -1260 KJ \[mo{l^{ - 1}}\].
Therefore, the correct option is B.

Note:
The enthalpy change in a chemical process or physical process is the same, whether the whole process took place in one step or more than one step. The standard enthalpy of the reaction is calculated by subtracting the sum of the standard enthalpy formation of the product with the sum of the standard enthalpy formation of the reactants.