Methanol is an organic solvent and is also used as fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol.
2CH3OH + 3O2 --> 2CO2 + 4H2O, (delta H = -1452.8 kJ/mol
just had this question on my webassign-
products
h2o is liquid: dH= -286 kJ (x4)
CO2 (g): dH= -393.509 kJ (x2)
reactants:
CH3OH (l): dH= 2dH (unknown)
O2 (g): dH= 0 (x3) (0 bc its elemental)
dH(rxn) = dH(prod) - dH(react)
-1452.8= -1931.018 - 2dH(CH3OH)
-2dH(CH3OH) = 478.218
dH(CH3OH) = -239.109
I don't agree with the answer provided by ash.
dHxn = (n*dHf products) - 2x = -1452.
Solve for x for dHf CH3OH.
Well, it's no wonder methanol is used as a fuel, it's quite the hotshot! Let's break down this fiery reaction, shall we?
First, we need to look at the balanced equation:
2 CH3OH + 3 O2 -> 2 CO2 + 4 H2O
The reaction tells us that when 2 moles of methanol react with 3 moles of oxygen gas, we end up with 2 moles of carbon dioxide and 4 moles of water.
Now, we know that the enthalpy change for this reaction, ΔH, is given as -1452.8 kJ/mol. This tells us the energy released or absorbed during the reaction.
Since we have the balanced equation, we can see that the stoichiometric coefficients for methanol, carbon dioxide, and water are 2, 2, and 4, respectively.
Using these coefficients, we can calculate the enthalpy change per mole of methanol:
(-1452.8 kJ/mol) / 2 = -726.4 kJ/mol
Therefore, the standard enthalpy of formation of methanol is -726.4 kJ/mol.
So, to sum it up, methanol not only has its hands all over organic solvents, but it's also bringing the heat as a fuel in automobile engines!
To calculate the standard enthalpy of formation of methanol, we need to use the given balanced chemical equation and apply Hess's Law. Hess's Law states that the enthalpy change for a reaction is the same whether it occurs in one step or a series of steps.
The standard enthalpy of formation (ΔHf°) of a substance is the enthalpy change when one mole of the substance is formed from its elements in their standard states at standard conditions (usually 25°C and 1 atm).
In this case, we have the balanced equation:
2CH3OH + 3O2 → 2CO2 + 4H2O (ΔH = -1452.8 kJ/mol)
The coefficients of this equation represent the stoichiometric ratio of the reactants and products, and the ΔH value represents the enthalpy change for the given reaction.
Now, let's break down the reaction into three steps:
Step 1: Formation of 2 moles of CO2
2CO2 → 2C + 4O2 (ΔH1)
Step 2: Formation of 4 moles of H2O
4H2O → 4H2 + 2O2 (ΔH2)
Step 3: Formation of 2 moles of CH3OH
2C + 4H2 + 2O2 → 2CH3OH (ΔH3)
We need to rearrange these steps to match the original equation. By reversing the steps, we can invert the signs of the enthalpy changes to obtain the desired reaction:
Step 3 (reversed): Formation of 2 moles of CH3OH
2CH3OH → 2C + 4H2 + 2O2 (-ΔH3)
Step 2 (reversed): Formation of 4 moles of H2O
4H2 + 2O2 → 4H2O (-ΔH2)
Step 1 (reversed): Formation of 2 moles of CO2
2C + 4O2 → 2CO2 (-ΔH1)
Now, we can sum up these equations to obtain the desired equation:
-ΔH1 + (-ΔH2) + (-ΔH3) = ΔHf°
Therefore, the standard enthalpy of formation of methanol (ΔHf°) can be calculated as:
ΔHf° = -ΔH1 + (-ΔH2) + (-ΔH3)
Substituting the given value, we have:
ΔHf° = -(-1452.8 kJ/mol) (since ΔH = -1452.8 kJ/mol)
So, the standard enthalpy of formation of methanol is 1452.8 kJ/mol.
(2*-393.5+4*-242)-(2*-293+0)
Delta H+ -1169 kJ/mol