1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298K and 1atm pressure according to the equation C+O2=CO2 during the reaction temperature rises from 298k to 299K. If the heat capacity of the bomb calorimeter is 20.7Kj/mole What is the enthalpy change for the above reaction.
Answers
Answer:
Enthalpy change (ΔH) will be 248.4 kJ/mole
Explanation:
Given data
pressure (P) = 1 atm
calorimeter's heat capacity (Cv) = 20.7 Kj/mole
change in temperature (ΔT) = -1 k (298 k -299 k)
- sign indicates that 1 k heat is evolved during this reaction
ΔH = ?
Solution
If q is the heat absorbed by the colorimeter , we can find out q by;
q = Cv × ΔT
q = 20.7 kJ/mole × 1 K
we do not put the negative sign as heat evolved by the reaction is absorbed by calorimeter
q = 20.7 kJ
It means that 1 g of graphite evolve 20 kJ heat during combustion, we want to now how much heat will evolve during combustion of one mole of graphite, as we know that one mole of graphite consists of 12 of graphite, so:
q = 20.7 kJ × 12 g ( 1 mole)
q = 248.4 kJ/mole
q is the total amount of heat of the system which is also called the enthalpy of the system
ΔH = 248.4 kJ/mole
Answer:
Enthalpy change =248.4kJ
