use thermochemical equations shown below to determine enthalpy for the reaction:
H2CO3-->H2O+CO2
1) H2CO3 --> H2CO + O2 H; 224 KJ
2)H2CO + O2 --> H2O + CO2 H; -100KJ
I know this problem is one that you have to rewrite the equations to get the equation in the original question - but I am lost on how to go about it. Thanks!!
To determine the enthalpy change for the overall reaction H2CO3 → H2O + CO2, you can use Hess's law. Hess's law states that the enthalpy change for a chemical reaction is independent of the pathway taken, as long as the initial and final conditions are the same.
To rewrite the given thermochemical equations and get the overall equation, you need to manipulate the equations so that the reactants and products of the desired reaction match.
Let's start with the given equations:
1) H2CO3 → H2CO + O2 ΔH = 224 kJ
2) H2CO + O2 → H2O + CO2 ΔH = -100 kJ
You need to multiply equation 1 by a factor to match the number of moles of H2CO. Looking at equation 2, we see that there is one mole of H2CO. So, multiply equation 1 by 1 to balance the number of moles of H2CO.
Now, the equations become:
1) H2CO3 → H2CO + O2 ΔH = 224 kJ
2) H2CO + O2 → H2O + CO2 ΔH = -100 kJ
Next, you need to ensure that the number of moles of O2 on the right side of equation 1 matches the number of moles of O2 on the left side of equation 2. Currently, there is one mole of O2 on both sides.
Now, the equations become:
1) H2CO3 → H2CO + O2 ΔH = 224 kJ
2) H2CO + O2 → H2O + CO2 ΔH = -100 kJ
Finally, you simply add the two equations together, canceling out the common species on each side:
1) H2CO3 → H2CO + O2 ΔH = 224 kJ
2) H2CO + O2 → H2O + CO2 ΔH = -100 kJ
---------------------------------------
H2CO3 → H2O + CO2 ΔH = 224 kJ - 100 kJ = 124 kJ
Therefore, the enthalpy change for the reaction H2CO3 → H2O + CO2 is 124 kJ.