Calculate H2
Equation 1- CH4g+2O2g+CO2g+2H2OgH1=-890kj
Equation 2- CH4g+O2g+CH2Og+H2OgH2=?
Equation 3- CH2Og++O2g+CO2g+H20gH3=-518kj
I don't know the question.
H2+H3=H1 look at equations
To calculate H2 in Equation 2, we need to use the data from Equations 1 and 3.
First, let's look at Equation 1:
CH4(g) + 2O2(g) + CO2(g) + 2H2O(g) H1 = -890 kJ
Here, we can see that 1 mole of CH4 reacts with 2 moles of O2 to yield 1 mole of CO2 and 2 moles of H2O. The enthalpy change for this reaction is -890 kJ.
Next, let's look at Equation 3:
CH2O(g) + O2(g) + CO2(g) + H2O(g) H3 = -518 kJ
Here, we can see that 1 mole of CH2O reacts with 1 mole of O2 to yield 1 mole of CO2 and 1 mole of H2O. The enthalpy change for this reaction is -518 kJ.
To calculate H2 in Equation 2, we can compare the reactants and products with the known equations.
Equation 2:
CH4(g) + O2(g) + CH2O(g) + H2O(g) H2 = ?
From Equation 1, we know that the combustion of 1 mole of CH4 produces 2 moles of H2O with an enthalpy change of -890 kJ.
From Equation 3, we know that the combustion of 1 mole of CH2O produces 1 mole of H2O with an enthalpy change of -518 kJ.
In Equation 2, we have 1 mole of CH4 and 1 mole of CH2O, which means the total mole of H2O produced is 2 + 1 = 3 moles.
The enthalpy change for the combustion of 3 moles of H2O will be the sum of the enthalpy changes in Equation 1 and Equation 3.
(-890 kJ) + (-518 kJ) = -1408 kJ
Therefore, the enthalpy change for Equation 2 (H2) is -1408 kJ.