Question

The enthalpies of formation of \( \text{CO}_2(g) \), \( \text{H}_2\text{O}(l) \), and \( \text{C}_6\text{H}_{12}\text{O}_6(s) \) are \( -393 \), \( -286 \), and \( -1170 \, \text{kJ/mol} \), respectively. The quantity of heat liberated when \( 18 \, \text{g} \) of \( \text{C}_6\text{H}_{12}\text{O}_6(s) \) is burnt completely in oxygen is.}

• A. $ 520 \, \text{kJ} $
• B. $ 145 \, \text{kJ} $
• C. $ 290 \, \text{kJ} $
• D. $ 420 \, \text{kJ} $

Hint:

To calculate the heat of combustion, use the standard enthalpies of formation and the stoichiometry of the balanced equation. Remember to account for the number of moles involved.

Answer 1

The Correct Option is C

Step 1: Known Information. 
Enthalpy of formation of \( \text{CO}_2(g) \): \( \Delta H_f^{\circ}(\text{CO}_2) = -393 \, \text{kJ/mol} \)
Enthalpy of formation of \( \text{H}_2\text{O}(l) \): \( \Delta H_f^{\circ}(\text{H}_2\text{O}) = -286 \, \text{kJ/mol} \)
Enthalpy of formation of \( \text{C}_6\text{H}_{12}\text{O}_6(s) \): \( \Delta H_f^{\circ}(\text{C}_6\text{H}_{12}\text{O}_6) = -1170 \, \text{kJ/mol} \)
Mass of \( \text{C}_6\text{H}_{12}\text{O}_6 \): \( 18 \, \text{g} \) 
Step 2: Balanced Combustion Equation. 
The balanced combustion equation for glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) is:
$$ \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) $$ Step 3: Calculate the Enthalpy Change for the Reaction. 
The standard enthalpy change of reaction (\( \Delta H_{\text{rxn}}^{\circ} \)) is given by:
$$ \Delta H_{\text{rxn}}^{\circ} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) $$ Reactants:
\( \text{C}_6\text{H}_{12}\text{O}_6(s) \): \( \Delta H_f^{\circ} = -1170 \, \text{kJ/mol} \)
\( \text{O}_2(g) \): \( \Delta H_f^{\circ} = 0 \, \text{kJ/mol} \) (standard state reference)
Products:
\( \text{CO}_2(g) \): \( \Delta H_f^{\circ} = -393 \, \text{kJ/mol} \)
\( \text{H}_2\text{O}(l) \): \( \Delta H_f^{\circ} = -286 \, \text{kJ/mol} \)
Substitute into the equation: $$ \Delta H_{\text{rxn}}^{\circ} = [6 \cdot (-393) + 6 \cdot (-286)] - [-1170 + 0] $$ Simplify: $$ \Delta H_{\text{rxn}}^{\circ} = [-2358 - 1716] - [-1170] $$ $$ \Delta H_{\text{rxn}}^{\circ} = -4074 + 1170 = -2904 \, \text{kJ/mol} $$ Step 4: Calculate Heat Liberated for 18 g of Glucose. 
The molar mass of glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) is: $$ 6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180 \, \text{g/mol} $$ Mass of glucose given: \( 18 \, \text{g} \)
Number of moles of glucose:
$$ \text{Moles} = \frac{18 \, \text{g}}{180 \, \text{g/mol}} = 0.1 \, \text{mol} $$ Heat liberated for \( 0.1 \, \text{mol} \):
$$ \text{Heat liberated} = 0.1 \times 2904 \, \text{kJ} = 290.4 \, \text{kJ} $$ Rounding to the nearest whole number: $$ \text{Heat liberated} = 290 \, \text{kJ} $$ Final Answer: \( \boxed{290 \, \text{kJ}} \)