Can someone please help guide me through this question?
Calculate the change in entropy of the surroundings . No work is done. You begin with 1 mol of CO. The reaction takes place at 298 K.
2CO + O2→ 2CO2
BDE (C≡O): 1072 kJ/mol; BDE (O=O): 495 kJ/mol; BDE (C=O): 799 kJ/mol
Sure! To calculate the change in entropy of the surroundings, we need to use the equation:
ΔS = ΔH / T
Where ΔS is the change in entropy, ΔH is the change in enthalpy, and T is the temperature in Kelvin.
In this case, since no work is done, the change in enthalpy (ΔH) can be calculated using the bond dissociation energies (BDEs) of the reactants and products.
The reaction you provided is 2CO + O2 → 2CO2.
To calculate the change in enthalpy, we need to consider the breaking of reactant bonds and the formation of product bonds.
Breaking bonds:
We have 2 moles of CO, and the bond dissociation energy (BDE) of C≡O is 1072 kJ/mol. So, the energy required to break the C≡O bond in 2 moles of CO is 2 * 1072 kJ/mol.
We also have 1 mole of O2, and the BDE of O=O is 495 kJ/mol. So, the energy required to break the O=O bond in 1 mole of O2 is 495 kJ/mol.
Formation of bonds:
We have 2 moles of CO2, and the BDE of C=O is 799 kJ/mol. So, the energy released when forming the C=O bond in 2 moles of CO2 is 2 * 799 kJ/mol.
Now, we can calculate the change in enthalpy (ΔH) by summing up the energy inputs and outputs:
ΔH = (Energy to break C≡O bonds + Energy to break O=O bonds) - (Energy released when forming C=O bonds)
ΔH = (2 * 1072 kJ/mol + 495 kJ/mol) - (2 * 799 kJ/mol)
Now, let's substitute the given values into the equation and calculate ΔH:
ΔH = (2 * 1072 kJ/mol + 495 kJ/mol) - (2 * 799 kJ/mol)
= 2642 kJ/mol - 1598 kJ/mol
= 1044 kJ/mol
Finally, we can use the formula ΔS = ΔH / T to calculate the change in entropy of the surroundings:
ΔS = ΔH / T
= 1044 kJ/mol / 298 K
≈ 3.5 kJ/(mol·K)
Therefore, the change in entropy of the surroundings is approximately 3.5 kJ/(mol·K).