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.

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.

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.

1.solid

2.carbon dioxide

3.helium