When 0.312g of glucose was burned in a bomb calorimeter (heat capacity 641 J/K) the temperature rose by 7.739 K. Calculate the standard enthalpy of combustion and the standard internal energy of combustion.
I've calculated the standard enthalpy but am unsure of how to go about finding the standard internal energy. Any suggestions?
I'm inclined to think that E = qv.
E = q+w and w = p*delta v. In a closed system, however, the p*delta v must be zero since delta v is zero and E = q.
1.5KJ
To calculate the standard enthalpy of combustion, you can use the formula:
ΔH = q/n
where ΔH is the change in enthalpy (in J/g), q is the heat absorbed (in J), and n is the amount of substance (in mol).
Given that the heat capacity of the bomb calorimeter is 641 J/K and the temperature rose by 7.739 K, you can calculate the heat absorbed by the bomb calorimeter using the formula:
q = C * ΔT
where C is the heat capacity (in J/K) and ΔT is the change in temperature (in K).
Plugging in the values, you get:
q = 641 J/K * 7.739 K = 4962.599 J
The amount of substance, n, can be calculated using the molar mass of glucose (C6H12O6).
molecular weight of glucose = 6 * atomic weight of Carbon + 12 * atomic weight of Hydrogen + 6 * atomic weight of Oxygen
= 6 * 12.01 g/mol + 12 * 1.01 g/mol + 6 * 16.00 g/mol
= 180.18 g/mol
Given that 0.312 g of glucose was burned, the number of moles of glucose is:
n = mass / molecular weight = 0.312 g / 180.18 g/mol = 0.001731 mol
Now, substituting the values back into the formula for ΔH:
ΔH = q/n = 4962.599 J / 0.001731 mol = 2864208.29 J/mol
Therefore, the standard enthalpy of combustion is 2,864,208.29 J/mol.
To find the standard internal energy of combustion, you can use the relation:
ΔE = ΔH - PΔV
where ΔE is the change in internal energy (in J), ΔH is the change in enthalpy (in J), P is the pressure (in atm), and ΔV is the change in volume (in L).
In the case of a bomb calorimeter, the volume change is assumed to be negligible since it is a closed system. Therefore, ΔV is considered to be zero.
Since the bomb calorimeter is at constant volume, the equation simplifies to:
ΔE = ΔH
So in this case, the standard internal energy of combustion is the same as the standard enthalpy of combustion.
Therefore, the standard internal energy of combustion is 2,864,208.29 J/mol.