Given the following heats of combustion.
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ
C(graphite) + O2(g) CO2(g) ΔH°rxn = -393.5 kJ
H2(g) + 1/2 O2(g) H2O(l) ΔH°rxn = -285.8 kJ
Calculate the enthalpy of formation of methanol (CH3OH) from its elements.
C(graphite) + 2 H2(g) + 1/2 O2(g) CH3OH(l)
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ
C(graphite) + O2(g) CO2(g) ΔH°rxn = -393.5 kJ
H2(g) + 1/2 O2(g) H2O(l) ΔH°rxn = -285.8 kJ
Reverse eqn 1, add in eqn 2, add in 2x the reverse of eqn 3. That will give you the final equation you want. When you reverse an eqn you must change the sign of delta H. If you multiply an equation, multiply delta H by the same number.
To calculate the enthalpy of formation of methanol (CH3OH) from its elements, we can use a Hess's law approach.
First, let's write the balanced equation for the formation of methanol from its elements:
C(graphite) + 2 H2(g) + 1/2 O2(g) → CH3OH(l)
Now, we can manipulate the given reactions in order to obtain the desired equation. We can reverse the third equation and multiply it by 2 to get the necessary number of moles:
2 H2O(l) → 2 H2(g) + O2(g) (reversed)
2 × (-285.8 kJ) → -571.6 kJ
Next, we need to reverse the second equation and multiply it by 2:
2 CO2(g) → 2 C(graphite) + 2 O2(g) (reversed)
2 × (-393.5 kJ) → -787.0 kJ
Finally, we sum up the manipulated equations to get the desired equation:
-787.0 kJ + -571.6 kJ + ΔH°rxn = -726.4 kJ
Now, let's solve for ΔH°rxn (enthalpy of formation of CH3OH):
ΔH°rxn = -726.4 kJ - (-787.0 kJ) - (-571.6 kJ)
= -726.4 kJ + 787.0 kJ - 571.6 kJ
= -510.8 kJ
Therefore, the enthalpy of formation of methanol (CH3OH) from its elements is -510.8 kJ.
To calculate the enthalpy of formation of methanol (CH3OH) from its elements, we need to use the given heats of combustion and Hess's Law.
Hess's Law states that the total enthalpy change of a reaction is independent of the pathway taken, and only depends on the initial and final states. In other words, if we can find a series of reactions where all the reactants and products cancel out except for the reaction we are interested in, we can sum up the enthalpy changes of those reactions to determine the enthalpy of formation of CH3OH.
Let's set up the reactions:
1. Multiply the second reaction by 2 to balance the carbon (C) atoms:
2C(graphite) + 2O2(g) -> 2CO2(g) (Multiply -393.5 kJ by 2)
2. Multiply the third reaction by 2 to balance the hydrogen (H) atoms:
2H2(g) + O2(g) -> 2H2O(l) (Multiply -285.8 kJ by 2)
3. Reverse the first reaction since we want to form CH3OH:
CO2(g) + 2H2O(l) -> CH3OH(l) + 3/2O2(g) (Change the sign of -726.4 kJ)
Now, we can add up these reactions to cancel out the common compounds, except for CH3OH:
2C(graphite) + 2O2(g) + 2H2(g) + O2(g) + CO2(g) + 2H2O(l) -> 2CO2(g) + 2H2O(l) + CH3OH(l) + 3/2O2(g)
Simplifying the equation, we get:
2C(graphite) + 3H2(g) + O2(g) + CO2(g) -> CH3OH(l)
Now, let's determine the enthalpy change of this overall reaction:
ΔH°rxn = ΔH°1 + ΔH°2 + ΔH°3
ΔH°rxn = (-393.5 kJ) + (2 * -285.8 kJ) + (-726.4 kJ)
ΔH°rxn = -393.5 kJ - 571.6 kJ - 726.4 kJ
ΔH°rxn = -1691.5 kJ
Therefore, the enthalpy of formation of methanol (CH3OH) from its elements is -1691.5 kJ.