Question

3.00g of glucose was burned in an excess of oxygen in bomb calorimeter with metal holder ("bomb") heat capacity of 2.21 kJ/oC. And 1.2kg of water where water has a specific heat capacity of 4.184 kJ/kgoC. The temp change upon combustion of glucose and oxygen was 19.0 0C to 25. 50C.Calculate the heat evolved from the combustion of 1.00 mol of glucose.

Answer 1

THE HEAT EVOLVED FROM THE COMBUSTION OF 1 MOLE OF GLUCOSE IS - 2819.7 KJ/MOL OR -2.82 * 10^3 KJ/MOL OF HEAT.

Step-by-step explanation:

Write out the variables given:

Mass of glucose = 3 g

Heat capacity of bomb calorimeter = 2.21 kJ /°C

Mass of water = 1.2 kg

Specific Heat capacity of water = 4.184 kJ/kg °C

Change in temperature = ( 25.50 °C - 19.0 °C ) = 6.5 °C

To calculate the heat evolved from the combustion of 1 mole of glucose, we do the following:

Equation for the reaction:

C6H1206 (s) + 6 02 (g)--------> 6 CO2 (g) + 6 H20 (l)

Calculate the total heat capacity involved in the system:

Heat capacity (Ctotal) = Heat capacity of the bomb calorimeter + heat capacity of water

Ctotal = 2.21 kJ/°C + (1.2 * 4.184 kJ/kg°C)

Ctotal = 7.2308 kJ°C

Next is to calculate the heat absoorbed by the calorimeter and water

Heat = Heat capacity (Ctotal) * change in temperature

Heat = 7.23 kJ/°C * 6.5 °C

Heat = 46.995 kJ

Hence, the amount of heat evolved when 3g of glucose was involved is 46.995 kJ

Since 1 mole of glucose reacts with 6 moles of oxygen to produce 6 moles of carbon dioxide and water respectively, then the amount of heat needed for the combustion of 1 mole of glucose is:

1 mole of glucose = (12 *6 + 1 * 12+ 16 *6) = 180 g/mol

From 3 g of glucose producing 46.995 kJ of heat

180 g of glucose will produce (180 * 46.995 / 3) kJ of heat

= 2819.7 kJ/mol of heat.

In conclusion, from the combustion of 1 mole of glucose, -2819.7 kJ/ mol of heat is evolved, since the heat was evolved or liberated.