Calculate the standard enthalpy change for the reaction 2C8H18(l)+17O2(g)> 16CO(g) + 18H2O(l)
Given
2C8H18(l)+25O2(g)>16CO2(g)+18H2O(l) deltaH=-11020 kj/mol
2CO(g)+ O2(g)>2CO2(g)
Delta H=-566.0 kj/mol
The > is supposed to be an arrow :)
delta Hrxn = (n*DHf products) - (n*DHf reactants)
To calculate the standard enthalpy change for the reaction, we can break it down into the sum of the individual reactions and then add them up.
1. Multiply the second reaction by 8 so that the coefficients of CO cancel out:
8 * (2CO(g) + O2(g) > 2CO2(g))
gives: 16CO(g) + 8O2(g) > 16CO2(g)
2. Multiply the first equation by 8/25 to match the coefficient of O2 in the second equation:
(2C8H18(l) + 25O2(g) > 16CO2(g) + 18H2O(l)) x (8/25)
gives: (16C8H18(l) + 8O2(g) > 128CO2(g) + 144H2O(l))
3. Now subtract the second equation from the modified first equation:
(16C8H18(l) + 8O2(g) > 128CO2(g) + 144H2O(l)) - (16CO(g) + 8O2(g) > 16CO2(g))
gives: (16C8H18(l) > 16CO(g) + 144H2O(l))
4. Lastly, add the modified second equation to the modified first equation:
16CO(g) + 144H2O(l) + 16CO2(g)
The new reaction is:
16C8H18(l) > 16CO(g) + 144H2O(l) + 16CO2(g)
Now sum up the enthalpy change values:
Delta H = -11020 kJ/mol + (-566.0 kJ/mol) = -11586 kJ/mol
So, the standard enthalpy change for the given reaction is -11586 kJ/mol.
To calculate the standard enthalpy change for the given reaction, we can use the concept of Hess's law. Hess's law states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states of the reaction.
First, we need to manipulate the given reactions to obtain the desired overall reaction:
Multiply the second reaction by 8 to balance the number of moles of CO:
16CO(g) + 8O2(g) -> 16CO2(g)
Delta H = -566.0 kj/mol
Multiply the first reaction by 8 and reverse it to balance the number of moles of CO2 and H2O:
16CO2(g) + 16H2O(l) -> 16CO(g) + 16O2(g)
Delta H = +11020 kj/mol
Now, we can add these two manipulated reactions to cancel out the intermediate species (16CO(g)):
8O2(g) + 16H2O(l) -> 16O2(g) + 8O2(g)
Delta H = (-566.0 kj/mol) + (+11020 kj/mol) = 10454.0 kj/mol
However, we need the reaction to be expressed in terms of 17 moles of O2, not 8 moles. We can use the stoichiometry of the reaction to relate the enthalpy change to 17 moles of O2:
Divide the enthalpy change by the coefficient of O2 in the original reaction (25 moles):
10454.0 kj/mol / 25 = 418.16 kj/mol
Therefore, the standard enthalpy change for the given reaction is 418.16 kJ/mol.