The standard enthalpy of combustion of C2H6O(l) is -1,367 kJ mol-1 at 298 K. What is the standard enthalpy of formation of C2H6O(l) at 298 K? Give your answer in kJ mol-1, rounded to the nearest kilojoule. Do not include units as part of your answer.
Note: The standard enthalpies of formation of CO2(g) and H2O(l) are -394 and -286 kJ mol-1, respectively, at 298 K.
C2H6O + 3O2 ==> 2CO2 + 3H2O
dHrxn = (n*dHfproducts) - (n*dHfreactants)
Substitute and solve for dHfC2H6O
Well, let's bring out the big comedy guns for this one!
Why did the chemist go broke? Because they couldn't find the enthalpy of formation in their pocket!
But seriously, calculating the enthalpy of formation involves using Hess's Law. So let's break it down:
The enthalpy of combustion of C2H6O(l) tells us how much energy is released when it is burned. In this case, it's -1,367 kJ mol-1.
To calculate the enthalpy of formation of C2H6O(l), we need to consider the reactants and products involved in the reaction and their enthalpies of formation.
C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
Now, using the enthalpies of formation for CO2(g) and H2O(l), we can solve for the enthalpy of formation of C2H6O(l).
So, (-1,367 kJ mol-1) = (2 * -394 kJ mol-1) + (3 * -286 kJ mol-1) + ΔHf
Simplifying the equation, we find:
ΔHf = (-1,367 kJ mol-1) - (2 * -394 kJ mol-1) - (3 * -286 kJ mol-1)
= -1,367 kJ mol-1 + 788 kJ mol-1 + 858 kJ mol-1
= 279 kJ mol-1
Rounded to the nearest kilojoule, the standard enthalpy of formation of C2H6O(l) at 298 K is 279 kJ mol-1.
There you have it, the answer we were all looking for! Let's hope it brings a smile to your face, or at least a grin. Remember, chemistry can always use a little humor to lighten things up!
To find the standard enthalpy of formation of C2H6O(l) at 298 K, we need to use the equation:
ΔH°f = ΣΔH°f(products) - ΣΔH°f(reactants)
In this case, the reactants are C2H6O(l), and the products are CO2(g) and H2O(l). The equation for the combustion reaction of C2H6O(l) can be written as:
C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
Using the given standard enthalpies of formation for CO2(g) and H2O(l), we can calculate the standard enthalpy of formation of C2H6O(l):
ΔH°f(C2H6O(l)) = ΣΔH°f(products) - ΣΔH°f(reactants)
= [2 * ΔH°f(CO2(g))] + [3 * ΔH°f(H2O(l))] - [ΔH°f(C2H6O(l))]
Substituting the values, we have:
ΔH°f(C2H6O(l)) = [2 * (-394 kJ/mol)] + [3 * (-286 kJ/mol)] - (-1,367 kJ/mol)
= -788 kJ/mol + (-858 kJ/mol) + 1,367 kJ/mol
= - :021 kJ/mol
Rounding to the nearest kilojoule, the standard enthalpy of formation of C2H6O(l) at 298 K is -2,023 kJ/mol.
To find the standard enthalpy of formation of C2H6O(l) at 298 K, we need to use the thermochemical equations and Hess's Law.
Hess's Law states that the overall enthalpy change of a reaction is independent of the reaction pathway and depends only on the initial and final states. We can use this concept to calculate the enthalpy of formation.
The reaction for the combustion of C2H6O(l) is:
C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
The enthalpy change for this combustion reaction is -1,367 kJ/mol. We need to manipulate this equation to match the given equation and use the enthalpies of formation of CO2(g) and H2O(l) to find the enthalpy of formation of C2H6O(l).
First, we need to reverse the equation by changing the sign of the enthalpy change to:
2CO2(g) + 3H2O(l) → C2H6O(l) + 3O2(g)
Next, we can apply Hess's Law to obtain the desired equation by subtracting the standard enthalpies of formation of CO2(g) and H2O(l):
ΔHf(C2H6O(l)) = ΣnΔHf(products) - ΣnΔHf(reactants)
where ΔHf is the enthalpy of formation, n is the stoichiometric coefficient, and the Σ symbol represents the sum of the enthalpies.
Using the given enthalpies of formation for CO2(g) (-394 kJ/mol) and H2O(l) (-286 kJ/mol), we can substitute these values into the equation:
ΔHf(C2H6O(l)) = 2(-394 kJ/mol) + 3(-286 kJ/mol)
ΔHf(C2H6O(l)) = -788 kJ/mol - 858 kJ/mol
ΔHf(C2H6O(l)) = -1646 kJ/mol
Rounding to the nearest kilojoule, the standard enthalpy of formation of C2H6O(l) at 298 K is -1646 kJ/mol.