"Synthesis gas" is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds. One reaction for producing synthesis gas is 3CH4 + 2H2O + CO2 = 4CO + 8H2
Using enthalpies of formation, calculate the enthalpy change for this reaction in kJ/mol.
Look up the delta Hf in your text or notes, then
delta Hrxn = (delta H products)-(delta H reactants). H2 will be zero. Therefore, it is
DHrxn = (4*DHof CO)-(3*DHCH4+2*DHH2O+DHCO2)
I ended up getting 747.5 kJ/mol
(4*-110.5) - (3*-74.81) + (2*-285.8) + (-393.5)
-442 + 1189.53 = 747.53 kJ/mol
You didn't write the states of the materials in the equation. I didn't look up any of the numbers but I recognize H2O as being in the liquid state. I'm wondering if this reaction you wrote isn't supposed to be gaseous state for H2O and/or at an elevated temperature (which would make it a gas). If it is you need to check all of your numbers to make sure you have them correct.
To calculate the enthalpy change for the given reaction, we need to use the enthalpies of formation of the reactants and products.
First, we need to know the enthalpies of formation of each compound listed in the balanced equation. Enthalpy of formation (ΔHf) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states at standard conditions (usually 25°C and 1 atm).
Using the given balanced equation: 3CH4 + 2H2O + CO2 = 4CO + 8H2
The enthalpy change for the reaction can be calculated using the following formula:
ΔHrxn = Σ(nΔHf products) - Σ(mΔHf reactants)
Where:
ΔHrxn = enthalpy change for the reaction
n = stoichiometric coefficient of the product
m = stoichiometric coefficient of the reactant
ΔHf = enthalpy of formation
Now, let's find the enthalpies of formation for each compound:
ΔHf(CH4) = -74.6 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
ΔHf(CO2) = -393.5 kJ/mol
ΔHf(CO) = -110.5 kJ/mol
ΔHf(H2) = 0 kJ/mol
Now we can substitute the values into the formula and calculate the enthalpy change:
ΔHrxn = [4ΔHf(CO)] + [8ΔHf(H2)] - [3ΔHf(CH4)] - [2ΔHf(H2O)] - [1ΔHf(CO2)]
ΔHrxn = [4(-110.5)] + [8(0)] - [3(-74.6)] - [2(-285.8)] - [1(-393.5)]
ΔHrxn = -442.0 + 0 + 223.8 + 571.6 + 393.5
ΔHrxn = 747.9 kJ/mol
Therefore, the enthalpy change for this reaction is 747.9 kJ/mol.