The combustion of methane in a laboratory burner is represented by the following equation:
CH4 + 2 O2 > CO2 + 2 H2O
Given the following three thermochemical equations:
C + H2 > CH4 H=-74.87kJ
C + O2 > CO2 H= -393.5
2 H2 + O2 > 2 H2O H=-571.6
a) Determine the molar heat of combustion of CH4.
b) Determine the enthalpy change when 6.4g of methane burns.
For A is it when I rearrange the three thermochemical equations so all products and reactants are corresponding to the equation on the very top. (I.E. CH4 in the first thermo chemical should be flipped so it corresponds with the top equation)
For B, I do not know how to figure out.
Those H values you have are for kJ/mol and the dH you calculated for the CH4 combustion is in kJ/mol. Let's call that value you obtained x (it would have been nice for you to let us know the value you obtained), then
convert 6.4 g CH4 to mols; i.e., 6.4/16 = ?
Then x value from above in kJ/mol x # mols in 6.4g = dH rxn for burning CH4.
I would have an answer for X, but I am confused who to approach a)
Nevermind I realized what I needed. Thanks!
So it would be
6.4/16.05=.40
then I would do .40(-890.23)=-354.98
Yes, I do these in one step.
(-890.3 kJ/16.05g) x 6.4 = ?
if -890.3 kJ/ was your calculated value for the combustion of CH4.
a) To determine the molar heat of combustion of CH4, we need to use Hess's Law, which states that the overall enthalpy change of a reaction is independent of the pathway taken. In this case, we want to find the enthalpy change for the combustion of methane.
From the given thermochemical equations, we can see that the combustion of methane is represented by the equation: CH4 + 2 O2 > CO2 + 2 H2O. We need to find the enthalpy change for this equation.
First, we need to manipulate the given equations so that the products and reactants correspond to the combustion equation. We can flip Equation 1 (C + H2 > CH4) to get CH4 > C + H2. We also multiply Equation 3 (2 H2 + O2 > 2 H2O) by 2 to get 4 H2 + 2 O2 > 4 H2O.
Now, we can add up these manipulated equations along with the given combustion equation, making sure to multiply the coefficients to balance the equation:
CH4 > C + H2
2 H2 + O2 > 2 H2O
C + O2 > CO2
------------------------
CH4 + 2 O2 > CO2 + 2 H2O
Next, we add up the enthalpy changes for each equation:
ΔH1 = -74.87 kJ (from equation 1)
ΔH2 = -571.6 kJ (from equation 3)
ΔH3 = -393.5 kJ (from equation 2, multiplied by 2)
Finally, we calculate the molar heat of combustion of CH4 by adding up the enthalpy changes:
ΔH(combustion) = ΔH1 + ΔH2 + ΔH3
= -74.87 kJ + (-571.6 kJ) + (-393.5 kJ × 2)
= -74.87 kJ - 571.6 kJ - 787 kJ
= -74.87 kJ - 1358.6 kJ
= -1433.47 kJ
Therefore, the molar heat of combustion of CH4 is -1433.47 kJ/mol.
b) To determine the enthalpy change when 6.4 g of methane (CH4) burns, we first need to calculate the moles of CH4 burned.
Molar mass of CH4 = 12.01 g/mol (C) + 4 * 1.008 g/mol (H) = 16.04 g/mol
Moles of CH4 = mass / molar mass = 6.4 g / 16.04 g/mol ≈ 0.3992 mol
Now, we can use the molar heat of combustion (ΔH(combustion)) from part a to calculate the enthalpy change:
Enthalpy change = ΔH(combustion) × moles of CH4
= -1433.47 kJ/mol × 0.3992 mol
= -572.2 kJ
Therefore, the enthalpy change when 6.4 g of methane burns is approximately -572.2 kJ.