Consider the following equations.
N2H4(l) + O2(g) N2(g) + 2 H2O(l) ÄH = -622.2 kJ
H2(g) + 1/2 O2(g) H2O(l) ÄH = -258.5 kJ
H2(g) + O2(g) H2O2(l) ÄH = -187.8 kJ
Use this information to calculate the enthalpy change for the reaction shown below.
N2H4(l) + 2 H2O2(l) N2(g) + 4 H2O(l) ÄH = ?
I got -764.kj. ia that correct?
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To calculate the enthalpy change for the given reaction, you can use the Hess's law of summation. This law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps.
The given equations are:
1. N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ΔH = -622.2 kJ
2. H2(g) + 1/2 O2(g) → H2O(l) ΔH = -258.5 kJ
3. H2(g) + O2(g) → H2O2(l) ΔH = -187.8 kJ
To obtain the desired reaction, you need to manipulate and combine the given equations. Here are the steps:
Step 1: Reverse equation 1:
N2(g) + 2 H2O(l) → N2H4(l) + O2(g) ΔH = +622.2 kJ
Step 2: Multiply equation 2 by 2:
2 H2(g) + O2(g) → 2 H2O(l) ΔH = -2(258.5) kJ = -517.0 kJ
Step 3: Multiply equation 3 by 2:
2 H2(g) + 2 O2(g) → 2 H2O2(l) ΔH = -2(187.8) kJ = -375.6 kJ
Step 4: Add the above three manipulated equations together:
N2(g) + 2 H2O(l) + 2 H2(g) + O2(g) + 2 H2(g) + 2 O2(g) → N2H4(l) + O2(g) + 2 H2O(l) + 2 H2O(l) + 2 H2O2(l)
Simplifying the equation:
N2(g) + 6 H2O(l) + 4 H2(g) + 4 O2(g) → N2H4(l) + O2(g) + 4 H2O(l) + 2 H2O2(l)
Now, add the ΔH values of each individual equation:
ΔHtotal = (622.2) + (-517.0) + (-375.6) = -270.4 kJ
Therefore, the enthalpy change for the reaction N2H4(l) + 2 H2O2(l) → N2(g) + 4 H2O(l) is -270.4 kJ, not -764 kJ.
To calculate the enthalpy change for the given reaction, let's use the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken between the reactants and the products. This means that we can use the given equations and their enthalpy changes to calculate the enthalpy change of the desired reaction.
First, let's rearrange the given equations and multiply them, if necessary, to match the desired reaction:
1. Equation 1 multiplied by 2:
2N2H4(l) + 4O2(g) → 2N2(g) + 4H2O(l) ΔH = -1244.4 kJ
2. Equation 2 multiplied by 2:
2H2(g) + O2(g) → 2H2O(l) ΔH = -517.0 kJ
Now, we can sum these equations to obtain the desired reaction:
2N2H4(l) + 4H2O2(l) → 2N2(g) + 4H2O(l) ΔH = ?
By adding the equations together, we get:
2N2H4(l) + 4H2O2(l) + 2H2(g) + O2(g) → 2N2(g) + 4H2O(l)
Now we can apply Hess's Law, which states that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.
Therefore, the enthalpy change of the desired reaction can be calculated as follows:
ΔH = [ΔH (equation 1)] + [ΔH (equation 2)]
ΔH = -1244.4 kJ + (-517.0 kJ)
ΔH = -1761.4 kJ
So, the enthalpy change for the given reaction is -1761.4 kJ, not -764 kJ as you calculated.
Please double-check your calculations to identify any potential errors.