Solveeit Logo
Login

Question

Question: The standard enthalpies of formation of \(C{{O}_{2}}(g)\), \({{H}_{2}}O(l)\) and glucose(s) at 25\(^...

The standard enthalpies of formation of CO2(g)C{{O}_{2}}(g), H2O(l){{H}_{2}}O(l) and glucose(s) at 250C^{0}C are -400kJmol−1kJmo{{l}^{-1}},-300 kJmol−1kJmo{{l}^{-1}} and -1300 kJmol−1kJmo{{l}^{-1}}, respectively. The standard enthalpy of combustion per gram of glucose at 25 0C^{0}Cis:
a.) +2900 kJ
b.) -2900 kJ
c.) -16.11 kJ
d.) +16.11 kJ

Explanation

Solution

To solve this question first we have to write all the equations involved in the formation of the given compound. The heat released when a sample is burned in the presence of air in its standard form to form a stable compound in its standard state is known as heat of combustion.

Complete step by step answer:
The equation involves in the combustion of glucose is:
C6H12O6(s)+CO2(g)→CO2(g)+H2O(l){{C}_{6}}{{H}_{12}}{{O}_{6}}(s)+C{{O}_{2}}(g)\to C{{O}_{2}}(g)+{{H}_{2}}O(l)

Balanced chemical equation is:
C6H12O6(s)+6CO2(g)→6CO2(g)+6H2O(l){{C}_{6}}{{H}_{12}}{{O}_{6}}(s)+6C{{O}_{2}}(g)\to 6C{{O}_{2}}(g)+6{{H}_{2}}O(l)

Given in the question:
The standard enthalpy of formation of CO2(g)C{{O}_{2}}(g)= -400 kJmol−1kJmo{{l}^{-1}}
The standard enthalpy of formation of H2O(l){{H}_{2}}O(l)= -300 kJmol−1kJmo{{l}^{-1}}
The standard enthalpy of formation of glucose(s) = -1300 kJmol−1kJmo{{l}^{-1}}

The change in enthalpy is calculated by subtracting the total change in enthalpy of product to the total change in enthalpy of the reactant. Which means that change in enthalpy will be equal to the sum of standard enthalpy of formation of water and the standard enthalpy of formation of CO2C{{O}_{2}} subtracting the standard enthalpy of formation of glucose.

$$$$\Delta {{H}^{0}}=(6 X (-400) + 6 X (-300)) - (-1300 + 0)$$ Hence the change in enthalpy = -2900 $kJmo{{l}^{-1}}$ Now, We know that the molecular weight of glucose = 180 g Change in enthalpy in $kJ{{g}^{-1}}$ will be equal to the ratio of change in enthalpy in $kJmo{{l}^{-1}}$ to the molecular weight in g. $$\Delta {{H}^{0}}$$ = $\dfrac{-2900}{180}$ = -16.11 Hence the standard enthalpy of combustion per gram of glucose is -16.11 kJ **So, the correct answer is “Option C”.** **Note:** Heat of formation is the amount of heat absorbed or evolved when 1 mole of a compound is formed from its constituent elements and each element is present in its normal physical state. The normal physical state can be solid, liquid or gases.