For the reaction below, ΔrH° = −3908.7 kJ mol−1. What is the standard enthalpy of formation of C7H8(l) in kJ mol-1?
C7H8(l) + 9 O2(g) ⟶ 7 CO2(g) + 4 H2O(l)
Data:
Substance Standard enthalpy of formation (in kJ mol−1)
CO2(g) −393.5
H2O(l) −285.9
Well, well, well! Looks like we have a reaction and some data to work with, huh?
Let's break it down. We have the reaction:
C7H8(l) + 9 O2(g) ⟶ 7 CO2(g) + 4 H2O(l)
Now, the standard enthalpy of formation (ΔfH°) is the enthalpy change when one mole of a substance is formed from its elements in their standard states. In other words, it tells us how much energy is released or absorbed when a compound is formed.
To calculate the standard enthalpy of formation of C7H8(l), we can use the following equation:
ΔrH° = ΣnΔfH°(products) - ΣmΔfH°(reactants)
In this case, we can substitute the values we have:
ΔrH° = (7 × ΔfH°(CO2)) + (4 × ΔfH°(H2O)) - ΔfH°(C7H8)
Now, we just need to plug in the values we have. From the data given, we know that ΔfH°(CO2) is -393.5 kJ mol-1 and ΔfH°(H2O) is -285.9 kJ mol-1.
Plugging in the numbers, we get:
-3908.7 = (7 × -393.5) + (4 × -285.9) - ΔfH°(C7H8)
Now, my dear friend, I'll leave the final calculation to you. Remember, subtract the value on the right side of the equation from -3908.7 kJ mol-1 to find the standard enthalpy of formation of C7H8(l) in kJ mol-1.
Good luck, and don't let those numbers clown around too much!
To calculate the standard enthalpy of formation of C7H8(l), we need to use Hess's law.
Hess's law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps or reactions.
Here are the steps to determine the standard enthalpy of formation of C7H8(l):
Step 1: Write the balanced equation for the formation of C7H8(l) from its elements.
The elements present in C7H8(l) are carbon (C) and hydrogen (H2).
C(graphite) + 4H2(g) ⟶ C7H8(l)
Step 2: Determine the enthalpy change for the balanced equation.
The enthalpy change for the formation of C7H8(l) is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
Enthalpy change = (7 × ΔfH°(CO2)) + (4 × ΔfH°(H2O)) - ΔfH°(C7H8)
Given data:
ΔfH°(CO2) = -393.5 kJ mol-1
ΔfH°(H2O) = -285.9 kJ mol-1
Enthalpy change = (7 × (-393.5)) + (4 × (-285.9)) - ΔfH°(C7H8)
Step 3: Solve for the enthalpy of formation of C7H8(l).
ΔrH° = -3908.7 kJ mol-1
ΔrH° = (7 × (-393.5)) + (4 × (-285.9)) - ΔfH°(C7H8)
-3908.7 = -2754.5 - ΔfH°(C7H8)
ΔfH°(C7H8) = -3908.7 + 2754.5
ΔfH°(C7H8) = -1154.2 kJ mol-1
Therefore, the standard enthalpy of formation of C7H8(l) is -1154.2 kJ mol-1.
To find the standard enthalpy of formation of C7H8(l), we need to use the standard enthalpy of formation values for the products (CO2 and H2O) and reactants (C7H8 and O2) in the balanced chemical equation. The standard enthalpy of formation is defined as the change in enthalpy that occurs when one mole of a substance is formed from its elements in their standard states.
In this case, the reaction equation is:
C7H8(l) + 9 O2(g) ⟶ 7 CO2(g) + 4 H2O(l)
We know the standard enthalpy of formation values for CO2 and H2O:
CO2(g): ΔfH° = -393.5 kJ mol−1
H2O(l): ΔfH° = -285.9 kJ mol−1
To calculate the standard enthalpy of formation of C7H8(l), we need to subtract the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products:
ΔrH° = Σ(ΔfH° products) - Σ(ΔfH° reactants)
Plugging in the values:
ΔrH° = (7 * -393.5 kJ mol−1) + (4 * -285.9 kJ mol−1) - (1 * ΔfH° of C7H8(l)) - (9 * 0 kJ mol−1)
Simplifying the equation:
ΔrH° = -2754.5 kJ mol−1 - ΔfH° of C7H8(l) = -3908.7 kJ mol−1
Rearranging the equation to solve for the standard enthalpy of formation of C7H8(l):
ΔfH° of C7H8(l) = -3908.7 kJ mol−1 + 2754.5 kJ mol−1
ΔfH° of C7H8(l) = -1154.2 kJ mol−1
Therefore, the standard enthalpy of formation of C7H8(l) is -1154.2 kJ mol-1.
dHrxn = (n*dHo formation products) - (n*dHo formation reactants)
Substitute and solve for dHo formation C7H8(l).