Question

Problem 6.6
lg of graphite is burnt in a bomb
calorimeter in excess of oxygen at 298 K
and 1 atmospheric pressure according to
the equation
C (graphite) + 0, (g) → CO, (g)
During the reaction, temperature rises
from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298K and 1 atm
​

Answer 1

Answer:

ressure (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

Explanation: