A 0.016076-mol sample of an organic compound was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 23.7oC to 24.5oC. If the heat capacity of the calorimeter is 4.98 kJ (oC)-1, then what is the constant volume heat of combustion of this compound, in kilojoules per mole? (Remember to include a "+" or "-" sign in your answer, as appropriate.)
Qv = Ccal x specific heat
That gives you Qv/0.01607 mol. Convert to 1 mol Add the appropriate sign. T increased meaning heat was emitted meaning dH is - and exothermic.
To find the constant volume heat of combustion of the compound, we can use the formula:
ΔH_combustion = q / n
where ΔH_combustion is the constant volume heat of combustion, q is the heat released by the combustion reaction, and n is the number of moles of the compound.
First, we need to calculate the heat released by the combustion reaction using the equation:
q = C * ΔT
where q is the heat released, C is the heat capacity of the calorimeter, and ΔT is the increase in temperature.
Given that the heat capacity of the calorimeter, C, is 4.98 kJ (oC)-1, and the increase in temperature, ΔT, is (24.5oC - 23.7oC) = 0.8oC, we can plug these values into the equation to calculate q:
q = C * ΔT
q = 4.98 kJ (oC)-1 * 0.8oC
q = 3.984 kJ
Now we have the heat released, q. Next, we need to find the number of moles of the compound, n.
Given that we have a 0.016076-mol sample of the compound, we can use this value for n.
Now, we can calculate the constant volume heat of combustion, ΔH_combustion, by dividing q by n:
ΔH_combustion = q / n
ΔH_combustion = 3.984 kJ / 0.016076 mol
ΔH_combustion ≈ 248.11 kJ/mol
Therefore, the constant volume heat of combustion of the compound is approximately 248.11 kJ/mol.