Question

use the bond enthalpies given below to estimate the enthalpy of reaction, in kj/mol, for the following reaction:
2 co(g) + o₂(g) → 2 co₂(g)

Explanation:

Step1: Calculate bond - breaking energy

For reactants: \(2\) moles of \(CO\) (\(C\equiv O\)) and \(1\) mole of \(O_2\) (\(O = O\)).
The bond - breaking energy \(E_{break}\):
For \(CO\): Each \(CO\) has \(1\) \(C\equiv O\) bond. For \(2\) \(CO\), the energy for breaking \(C\equiv O\) bonds is \(2\times1070\) kJ/mol.
For \(O_2\): Each \(O_2\) has \(1\) \(O = O\) bond. The energy for breaking \(O = O\) bond is \(498.7\) kJ/mol.
So, \(E_{break}=2\times1070 + 498.7=2140+498.7 = 2638.7\) kJ/mol.

Step2: Calculate bond - forming energy

For products: \(2\) moles of \(CO_2\) (\(O = C = O\)). Each \(CO_2\) has \(2\) \(C = O\) bonds.
For \(2\) \(CO_2\), the number of \(C = O\) bonds is \(2\times2 = 4\).
The bond - forming energy \(E_{form}=4\times799=3196\) kJ/mol.

Step3: Calculate the enthalpy of reaction

The enthalpy of reaction \(\Delta H=E_{break}-E_{form}\).
\(\Delta H = 2638.7-3196=- 557.3\) kJ/mol.

Answer:

\(-557.3\) kJ/mol