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!!
H2CO3>>H2CO+O2>>H2O+CO2
Hr= 224+(-100) Kj/mole
To determine the enthalpy change (ΔH) for the reaction: H2CO3 -> H2O + CO2, we can rearrange and combine the given thermochemical equations.
Here's how you can do it step-by-step:
Step 1: Reverse Equation 2
H2CO + O2 -> H2O + CO2 (reversed) ΔH = +100 KJ
Step 2: Multiply Equation 1 by 2
2(H2CO3 -> H2CO + O2) ΔH = 2(224 KJ) = +448 KJ
Step 3: Cancel out the H2CO term
H2CO3 -> H2CO + O2 (Equation 1)
H2CO + O2 -> H2O + CO2 (reversed of Equation 2)
Step 4: Add both equations together
H2CO3 -> H2CO + O2
+ H2CO + O2 -> H2O + CO2 (reversed)
H2CO3 -> H2O + CO2
Step 5: Find the net enthalpy change
ΔH_net = ΔH1 + ΔH2
ΔH_net = +448 KJ + (+100 KJ)
ΔH_net = +548 KJ
Therefore, the enthalpy change (ΔH) for the reaction H2CO3 -> H2O + CO2 is +548 KJ.
To determine the enthalpy for the given reaction, you need to manipulate the given thermochemical equations so that they can be added up to obtain the desired overall reaction.
Let's start by analyzing the given equations:
1) H2CO3 --> H2CO + O2 ΔH = 224 kJ
2) H2CO + O2 --> H2O + CO2 ΔH = -100 kJ
To obtain the overall reaction H2CO3 --> H2O + CO2, we need to cancel out the common compounds on the left and right sides of the equations.
1) H2CO3 --> H2CO + O2
Since we have H2CO3 on the left side and H2CO on the right side, we can cancel out H2CO by multiplying the first equation by 1 and the second equation by 1. This will allow the H2CO compound to cancel out.
1 × [H2CO3 --> H2CO + O2]
1 × [H2CO + O2 --> H2O + CO2]
Now, let's add the two equations together to get the overall reaction:
H2CO3 + H2CO + O2 --> H2O + CO2 + O2
However, we still have an O2 molecule on the left side that is not present on the right side of the overall reaction. To balance this out, we need to include the O2 molecule on the right side as well. By doing that, the overall reaction becomes:
H2CO3 + H2CO + O2 --> H2O + CO2 + O2
Now, we can sum up the enthalpy changes of the individual equations to obtain the enthalpy change of the overall reaction:
ΔH = ΔH1 + ΔH2
= 224 kJ + (-100 kJ)
= 124 kJ
Therefore, the enthalpy change for the given reaction H2CO3 --> H2O + CO2 is 124 kJ.