The standard enthalpy of formation at 298k for ccl4,h2o,co2,hcl are -22.5,-57.8,-94.1 and-22.1 kcal/mol.calculate ▲rH FOR reaction ccl4+2h2o→co2+HCL
dHrxn = (n*dHformation products) - (n*dHformation reactants)
-41.4Kcal
To calculate the change in enthalpy (▲rH) for the given reaction, we need to use Hess's Law, which states that the change in enthalpy for a reaction is equal to the sum of the enthalpies of formation for the products minus the sum of the enthalpies of formation for the reactants.
The given reaction is:
CCl4 + 2H2O → CO2 + HCl
The enthalpy change (▲rH) can be calculated by using the enthalpies of formation for the products and reactants. The enthalpy of formation for each compound is given:
∆Hf (CCl4) = -22.5 kcal/mol
∆Hf (H2O) = -57.8 kcal/mol
∆Hf (CO2) = -94.1 kcal/mol
∆Hf (HCl) = -22.1 kcal/mol
Now let's use the equation for ▲rH:
▲rH = ∑∆Hf (products) - ∑∆Hf (reactants)
For the given reaction, the products are CO2 and HCl, and the reactants are CCl4 and H2O.
▲rH = [∆Hf (CO2) + ∆Hf (HCl)] - [∆Hf (CCl4) + 2∆Hf (H2O)]
▲rH = [(-94.1 kcal/mol) + (-22.1 kcal/mol)] - [(-22.5 kcal/mol) + 2(-57.8 kcal/mol)]
▲rH = -116.2 kcal/mol - (-138.1 kcal/mol)
▲rH = -116.2 kcal/mol + 138.1 kcal/mol
▲rH = 21.9 kcal/mol
Therefore, the change in enthalpy (▲rH) for the given reaction is 21.9 kcal/mol.