A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents?

The correct answer is 15.54 kJ per degree C.

Explanation:

The enthalpy change for one mole of a substance, which combines or burns with the oxygen under the standard conditions, that is, at 25 degree C and 1 bar pressure is known as the standard molar enthalpy of combustion. The amount of heat transferred can be calculated by using the formula, q = mcΔT (i)

Here q is the amount of heat transferred, c is the specific heat, ΔT is the change in temperature, and m is the mass of the substance. As in case of bomb calorimeter, mass if considered constant, thus, for calorimeter the equation mentioned will become, q = cΔT (ii)

The standard molar enthalpy of combustion for carbon is -393.5 kJ/mol, that is, -393.5 kJ per mole of heat is generated by burning one mole of carbon. The molecular mass of carbon is 12 gram per mole.  

Thus, the number of moles of carbon equivalent to 0.562 grams of carbon can be determined as,  

Number of moles of carbon = mass / molecular mas

= 0.562 grams / 12 gram per mole

= 0.047 mol

The heat generated by burning 0.562 grams or 0.047 mole will be,  

q = ΔH° × number of moles

= (-393.51 kJ/mol) × 0.047 mol

= -18.49 kJ, the negative sign shows that the heat is produced.  

To find heat capacity of calorimeter, put the value of q as -18.49 kJ, for ΔT as (27.93 °C - 26.74 °C) in the equation (ii)

18.49 kJ = c × (27.93 - 26.74)

c = 18.49 kJ/1.19 °C

c = 15.54 kJ/°C

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

b.final