Complete the Hess’ cycle using the values given in the table and hence calculate the enthalpy
change, ∆Hr
, for this reaction.
Show your working.
substance ∆Hf / kJmol–1
NO2(g) 34.0
H2O(l) –286
HNO3(l) –173
NO(g) 91.1
∆Hr
3NO2(g) + H2O(l) 2HNO3(l) + NO(g)
dHrxn = (n*dHoproducts) - (n*dHoreactants)
Substitute the values and calculate. Show your work if you get stuck.
Well, it seems like you're challenging me with some chemistry, a subject I find quite electrifying! Let's dive into the world of Hess' cycle and calculate the enthalpy change (∆Hr).
To complete the Hess' cycle, we need to break down the overall reaction into a series of intermediate reactions. Let's start with the given substances and their enthalpies of formation (∆Hf):
1. The formation of 2HNO3(l):
NO2(g) + 0.5O2(g) → HNO3(l)
The enthalpy change for this reaction is zero since it involves the formation of a compound in its standard state.
2. The decomposition of H2O(l):
H2O(l) → H2(g) + 0.5O2(g)
Since we don't have the enthalpy change for this reaction directly, we can use the enthalpy change for the reverse reaction, which is the formation of water:
H2(g) + 0.5O2(g) → H2O(l)
The enthalpy change for the reverse reaction is -286 kJ/mol (given) which means the enthalpy change for the decomposition of water is +286 kJ/mol.
3. The formation of NO(g) from the elements:
0.5N2(g) + 0.5O2(g) → NO(g)
Again, we don't have the enthalpy change for this reaction directly, but we can use the enthalpy change for the formation of nitrogen dioxide (NO2):
N2(g) + 2O2(g) → 2NO2(g)
The enthalpy change for the formation of NO2(g) is 2 * 34.0 kJ/mol = +68 kJ/mol. Therefore, the enthalpy change for the formation of NO(g) is -68 kJ/mol.
Now, let's add up the enthalpy changes for the intermediate reactions:
2HNO3(l) + NO(g) - 3NO2(g) - H2O(l)
Since enthalpy change is a state function, we can simply sum up the enthalpy changes of the individual reactions.
∆Hr = 0 kJ/mol + (–68 kJ/mol) + (–286 kJ/mol) = -354 kJ/mol
Therefore, the enthalpy change (∆Hr) for the given reaction is -354 kJ/mol. And I must say, the path to finding this value was a-mole-zing!
To complete the Hess' cycle and calculate the enthalpy change (∆Hr) for the given reaction, we need to break down the reaction into a series of intermediate steps, and then sum up the enthalpy changes for each step.
Step 1: Formation of reactant NO2(g)
2NO(g) + O2(g) → 2NO2(g)
Given: ∆Hf(NO2) = 34.0 kJ/mol
Given: ∆Hf(NO) = 91.1 kJ/mol
Given: ∆Hf(O2) = 0 kJ/mol (Oxygen is in its standard state)
Since 2NO(g) is a reactant in the reaction, we need to reverse the reaction and multiply the enthalpy change (∆Hf) by -1:
2NO2(g) → 2NO(g) + O2(g)
∆H1 = -2 * ∆Hf(NO2) = -2 * 34.0 kJ/mol = -68.0 kJ/mol
Step 2: Formation of H2O(l)
Given: ∆Hf(H2O) = -286 kJ/mol
Step 3: Formation of product HNO3(l)
Given: ∆Hf(HNO3) = -173 kJ/mol
Step 4: Formation of product NO(g)
Given: ∆Hf(NO) = 91.1 kJ/mol
Step 5: Summing up the enthalpy changes
Finally, to determine the enthalpy change (∆Hr) for the reaction:
∆Hr = ∆H1 + ∆Hf(H2O) + 2∆Hf(HNO3) + ∆Hf(NO)
∆Hr = -68.0 kJ/mol + (-286 kJ/mol) + 2*(-173 kJ/mol) + 91.1 kJ/mol
Calculating the value:
∆Hr = -68.0 kJ/mol - 286 kJ/mol - 346 kJ/mol + 91.1 kJ/mol
∆Hr = -609 kJ/mol + 91.1 kJ/mol
∆Hr = -517.9 kJ/mol
Therefore, the enthalpy change (∆Hr) for the given reaction is -517.9 kJ/mol.
To complete the Hess' cycle and calculate the enthalpy change (∆Hr) for the given reaction, we'll need to break down the reaction into smaller steps and use the known enthalpy changes for those steps.
The given reaction is:
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
To complete the cycle, we need to decompose or combine known reactions to obtain the overall reaction.
Step 1: Decompose HNO3(l):
HNO3(l) → NO2(g) + O2(g)
This step can be written by reversing the ΔHf value of NO2(g):
ΔH1 = -ΔHf(NO2(g)) = -34.0 kJ/mol
Step 2: Decompose NO(g):
NO(g) → 1/2 N2(g) + 1/2 O2(g)
This step can be written by combining the ΔHf values of N2(g) and O2(g):
ΔH2 = ΔHf(N2(g)) + ΔHf(O2(g)) = 0 kJ/mol + 0 kJ/mol = 0 kJ/mol
Step 3: Combine steps 1 and 2, and multiply them by 2 to obtain the desired equation:
2HNO3(l) + NO(g) → 2NO2(g) + O2(g)
Add the enthalpy changes of both steps to get the overall enthalpy change (∆Hr):
∆Hr = 2ΔH1 + ΔH2
Substitute the known values into the equation:
∆Hr = 2(-34.0 kJ/mol) + 0 kJ/mol
∆Hr = -68.0 kJ/mol
Therefore, the enthalpy change (∆Hr) for the given reaction is -68.0 kJ/mol.