Question

Assume that 0.491 g of diborane is combusted in a calorimeter whose heat capacity (Ccalorimeter) is 7.854 kJ/°C at 19.63°C. What is the final temperature of the calorimeter?

ΔHcombustion for this process is –1958 kJ/mol.

Answer 1

To solve this problem, we can use the formula:

q = -Ccalorimeter x ΔT

where q is the heat absorbed by the calorimeter, Ccalorimeter is the heat capacity of the calorimeter, and ΔT is the change in temperature of the calorimeter.

First, we need to calculate the amount of heat released by the combustion of diborane. We can use the molar mass of diborane to convert the given mass to moles:

moles of diborane = 0.491 g / 27.66 g/mol = 0.01775 mol

The heat released by the combustion of 1 mole of diborane is -1958 kJ, so the heat released by the combustion of 0.01775 mol is:

q = 0.01775 mol x (-1958 kJ/mol) = -34.76 kJ

The negative sign indicates that heat is released by the reaction.

Now we can use the formula above to find the change in temperature of the calorimeter:

-34.76 kJ = -7.854 kJ/°C x ΔT

ΔT = 4.43°C

Therefore, the final temperature of the calorimeter is 19.63°C - 4.43°C = 15.20°C.