Question

Consider the reaction C12H22O11(s)+12O2(g)→12CO2(g)+11H2O(l) in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/∘C. The temperature increase inside the calorimeter was found to be 22.0 ∘C. Calculate the change in internal energy, ΔE, for this reaction per mole of sucrose. Express the change in internal energy in kilojoules per mole to three significant figures.

Answer 1

Answer : The internal energy change for 1 mole of sucrose is $-5.65\times 10^3kJ/mole$

Explanation :

First we have to calculate the heat released by calorimeter.

$q_{calorimeter}=-c\times \Delta T$

where,

$q_{calorimeter}$ = heat released by calorimeter = ?

c = heat capacity of bomb calorimeter = $7.50kJ/^oC$

$\Delta T$ = change in temperature of calorimeter = $22.0^oC$

Now put all the given values in the above formula, we get

$q_{calorimeter}=-7.50kJ/^oC\times 22.0^oC$

$q_{calorimeter}=-165kJ$

Now we have to calculate the moles of sucrose.

$\text{Moles of sucrose}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}}=\frac{10.0g}{342.3g/mol}=0.0292mol$

Now we have to calculate the internal energy change.

As we know that, bomb calorimeter carried out at constant volume condition.

$\Delta q=\Delta E+P\Delta V$

here, $\Delta V=0$

So,

$\Delta q=\Delta E=-165kJ$

Now we have to calculate the internal energy change for 1 mole of sucrose.

$\Delta E=\frac{\Delta q}{n}=\frac{-165kJ}{0.0292mole}=-5650.7kJ/mole=-5.65\times 10^3kJ/mole$

Therefore, the internal energy change for 1 mole of sucrose is $-5.65\times 10^3kJ/mole$