Question

1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298K and 1atm pressure according to the equation
$C_{(graphite)} +O_{2 (g)} ----\ \textgreater \ CO_{2(g)}$

During the reaction, temperature rises from 298K to 299K. If the heat capacity of the bomb calorimeter is $20.7 kJ/K$, what is the enthalpy change for the above reaction at 298K and 1atm?

(1) $20.7 kJ mol^{-1}$

(2) $-20.7 kJmol^{-1}$

(3) $-248 kJmol^{-1}$

(4) $248 kJmol^{-1}$

Answer 1

Answer:

Explanation:

C (graphite) + O2 (g) → CO2

Option (3) = -284kJmol-1