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): ∆Hº = -11,020 kJ/mol
2CO(g) + O2(g) --> 2CO2(g): ∆Hº = -566.0 kJ/mol
To find the standard enthalpy change for the given reaction, we will use the Hess's Law, which states that the enthalpy change for a reaction is the sum of the enthalpy changes of the individual steps to convert reactants to products.
Step 1: Multiply the second reaction by 8 to balance the carbon dioxide (CO2) on both sides:
16CO(g) + 8O2(g) --> 16CO2(g)
Step 2: Multiply the third reaction by 16 to balance the carbon monoxide (CO) on both sides:
32CO(g) + 16O2(g) --> 32CO2(g)
Step 3: Combine the three equations, canceling out common species:
2C8H18(l) + 17O2(g) --> 16CO(g) + 18H2O(l)
16CO(g) + 8O2(g) --> 16CO2(g)
32CO(g) + 16O2(g) --> 32CO2(g)
The canceled species are 16CO(g), 8O2(g), and 16CO2(g), leaving us with:
2C8H18(l) + 17O2(g) --> 18H2O(l) + 32CO(g)
Now, add up the enthalpy changes of the individual reactions:
∆Hº(overall) = ∆Hº(reaction 1) + ∆Hº(reaction 2) + ∆Hº(reaction 3)
∆Hº(overall) = 2 * ∆Hº(reaction 1) + 1 * ∆Hº(reaction 2) + 1 * ∆Hº(reaction 3)
∆Hº(overall) = 2 * (-11,020 kJ/mol) + (-566.0 kJ/mol) + 0 kJ/mol
∆Hº(overall) = -22,040 kJ/mol - 566.0 kJ/mol
∆Hº(overall) = -22,606 kJ/mol
Therefore, the standard enthalpy change (∆Hº) for the given reaction is -22,606 kJ/mol.
To calculate the standard enthalpy change for the reaction:
2C8H18(l) + 17O2(g) --> 16CO(g) + 18H2O(l)
You need to use Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.
1. First, note that the given reactions have different stoichiometry than the target reaction, so we need to manipulate the given reactions to match the target reaction stoichiometry.
a. Multiply the second given reaction by 8 to make the number of moles of CO2 created equal to the number of moles of CO required in the target reaction:
8CO(g) + 8O2(g) --> 8CO2(g): ∆Hº = -4,528.0 kJ/mol
b. Multiply the first given reaction by 2 to make the number of moles of CO2 produced equal to the number of moles of CO in the above equation:
4C8H18(l) + 50O2(g) --> 32CO2(g) + 36H2O(l): ∆Hº = -22,040 kJ/mol
2. Now, add the two manipulated reactions together to obtain the target reaction:
4C8H18(l) + 50O2(g) + 8CO(g) + 8O2(g) --> 32CO2(g) + 36H2O(l) + 8CO(g)
Cancel out any common species on both sides:
4C8H18(l) + 42O2(g) --> 32CO2(g) + 36H2O(l)
3. Now, subtract the given reaction that has CO(g) and CO2(g) as products to obtain the target reaction:
(4C8H18(l) + 42O2(g) --> 32CO2(g) + 36H2O(l)) - (4C8H18(l) + 50O2(g) --> 32CO2(g) + 36H2O(l))
The result is:
8O2(g) --> 0
4. Since O2(g) is a diatomic gas, it is considered to be in its standard state, so its enthalpy change is zero:
∆Hº = 0 kJ/mol
Therefore, the standard enthalpy change for the reaction:
2C8H18(l) + 17O2(g) --> 16CO(g) + 18H2O(l)
is 0 kJ/mol.