Given the data in the table below, ∆Horxn for the reaction
3NO2(g) + H2O(l) → 2HNO3(aq) + 3NO(g)
is __________ kJ.
Substance
∆Horxn (kJ/mol)
H2O(l)= -286
NO(g)= 90.0
NO2(g)= 34.0
HNO3(aq)= -207
NH3(g)= -46.0
A) -140
B) -508
C) -64
D) 140
E) 64
Well, it's important to remember that ΔHorxn represents the change in enthalpy for a reaction. To calculate ΔHorxn, we need to sum up the enthalpies of the products and subtract the enthalpies of the reactants.
In this case, we have:
Reactants:
3NO2(g) = 3 * 34.0 kJ/mol = 102.0 kJ/mol
H2O(l) = -286 kJ/mol
Products:
2HNO3(aq) = 2 * -207 kJ/mol = -414 kJ/mol
3NO(g) = 3 * 90.0 kJ/mol = 270 kJ/mol
Now, let's calculate the overall enthalpy change:
∆Horxn = (sum of products) - (sum of reactants)
= (-414 kJ/mol + 270 kJ/mol) - (102.0 kJ/mol - 286 kJ/mol)
= -144 kJ/mol + 388 kJ/mol
= 244 kJ/mol
So, the correct answer is D) 244 kJ/mol.
But hey, don't worry if you got it wrong. Chemistry can be a lot to handle sometimes. Just remember, it's all about the reactions!
To calculate the ∆Horxn for the reaction, we need to calculate the sum of the ∆Horxn for the products minus the sum of the ∆Horxn for the reactants.
Reactants:
3NO2(g) + H2O(l)
Products:
2HNO3(aq) + 3NO(g)
∆Horxn = (∆Horxn of products) - (∆Horxn of reactants)
∆Horxn = (2HNO3(aq) + 3NO(g)) - (3NO2(g) + H2O(l))
To calculate the sum of the ∆Horxn for the products:
(2 * ∆Horxn of HNO3(aq)) + (3 * ∆Horxn of NO(g))
= (2 * -207) + (3 * 90)
= -414 + 270
= -144 kJ
To calculate the sum of the ∆Horxn for the reactants:
(3 * ∆Horxn of NO2(g)) + (∆Horxn of H2O(l))
= (3 * 34) + (-286)
= 102 - 286
= -184 kJ
Now, plug these values into the equation:
∆Horxn = (-144) - (-184)
= -144 + 184
= 40 kJ
The ∆Horxn for the given reaction is 40 kJ.
Therefore, the correct answer is not listed.
To find the enthalpy change (∆Horxn) for the given reaction, you need to calculate the overall sum of the enthalpies of the products minus the overall sum of the enthalpies of the reactants.
Using the given data, we can determine the enthalpy changes for the reactants and products:
Reactants:
3NO2(g): 3 * ∆Horxn(NO2(g)) = 3 * 34.0 kJ/mol
H2O(l): ∆Horxn(H2O(l)) = -286.0 kJ/mol
Products:
2HNO3(aq): 2 * ∆Horxn(HNO3(aq)) = 2 * (-207.0) kJ/mol
3NO(g): 3 * ∆Horxn(NO(g)) = 3 * 90.0 kJ/mol
Now, add up the enthalpy changes for the reactants and products:
Reactants: 3 * 34.0 kJ/mol + (-286.0 kJ/mol)
Products: 2 * (-207.0 kJ/mol) + 3 * 90.0 kJ/mol
Simplifying the expressions:
Reactants: 102.0 kJ/mol - 286.0 kJ/mol = -184.0 kJ/mol
Products: -414.0 kJ/mol + 270.0 kJ/mol = -144.0 kJ/mol
Finally, calculate the overall ∆Horxn = ∆Horxn(Products) - ∆Horxn(Reactants):
∆Horxn = -144.0 kJ/mol - (-184.0 kJ/mol)
∆Horxn = -144.0 kJ/mol + 184.0 kJ/mol
∆Horxn = 40.0 kJ/mol
Therefore, the correct answer is not provided in the options.