Calculate (delta) H rxn for
2NOCl(g)---- N2(g) + O2(g) + Cl(g)
given the following set of reactions:
1/2N2(g) + 1/2O2(g)-----NO(g) delta H =90.3kJ
NO(g) + 1/2Cl2(g)----NOCl(g) delta H = -38.6 kJ
Are you sure that equation 1 is correct? It isn't balanced. I think it should be
2NOCl ==> N2 + O2 + Cl2
To get that equation, multiply equation 2 by 2 and reverse it (multiply delta H by 2 and change the sign). Multiply equation 3 by 2 and reverse it (multiply delta H by 2 and change the sign). Then add the two equations to obtain the final equation. Add the multiplied delta Hs with the changed signs.
Well, to calculate the overall ΔH for the reaction, we can use Hess's Law. We'll start by writing the given reactions in the reverse direction along with their corresponding ΔH values:
NO(g) + 1/2Cl2(g) → NOCl(g) ΔH = +38.6 kJ
1/2N2(g) + 1/2O2(g) → NO(g) ΔH = -90.3 kJ
Now, let's add both equations together and cancel out any common species:
NO(g) + 1/2Cl2(g) + 1/2N2(g) + 1/2O2(g) → NOCl(g) + NO(g)
By canceling out NO(g), we get the balanced equation:
1/2Cl2(g) + 1/2N2(g) + 1/2O2(g) → NOCl(g)
Now, let's add the ΔH values for the reactions:
ΔH for the overall reaction = ΔH1 + ΔH2
ΔH for the overall reaction = +38.6 kJ + (-90.3 kJ)
ΔH for the overall reaction = -51.7 kJ
So, the ΔH for the reaction 2NOCl(g) → N2(g) + O2(g) + Cl(g) is -51.7 kJ. But hey, don't be negative about it! Every reaction needs a little spice!
To calculate ΔH_rxn for the reaction 2NOCl(g) -> N2(g) + O2(g) + Cl(g), we can use the given set of reactions and their respective enthalpy changes (ΔH).
First, let's write the target reaction in terms of the given reactions:
2NOCl(g) -> 2(NO(g) + 1/2Cl2(g))
or
2NOCl(g) -> 2NO(g) + Cl2(g)
We can now sum up the reactions together and cancel out the common substances:
(1/2N2(g) + 1/2O2(g)) + (NO(g) + 1/2Cl2(g)) -> 2(NO(g) + 1/2Cl2(g)) + N2(g) + O2(g)
or
1/2N2(g) + 1/2O2(g) + NO(g) + 1/2Cl2(g) -> 2NO(g) + Cl2(g) + N2(g) + O2(g)
Now, let's calculate the overall change in enthalpy (ΔH_rxn) by summing up the enthalpy changes of the individual reactions:
ΔH_rxn = (1/2N2(g) + 1/2O2(g)) + (NO(g) + 1/2Cl2(g)) - (2NO(g) + Cl2(g) + N2(g) + O2(g))
Since the ΔH values are given as follows:
ΔH1 = 90.3 kJ (for 1/2N2(g) + 1/2O2(g) -> NO(g))
ΔH2 = -38.6 kJ (for NO(g) + 1/2Cl2(g) -> NOCl(g))
Substituting these values:
ΔH_rxn = (90.3 kJ) + (-38.6 kJ) - (2NO(g) + Cl2(g) + N2(g) + O2(g))
However, we still need the values for the enthalpies of formation of NO(g), Cl2(g), N2(g), and O2(g) to calculate the overall ΔH_rxn. Unfortunately, these values are not provided in the given information.
Hence, without the enthalpies of formation of the individual species, we cannot calculate the exact value of ΔH_rxn for the given reaction.
To calculate ΔH_rxn for the reaction 2NOCl(g) → N2(g) + O2(g) + Cl(g) using the given set of reactions, you can use the concept of Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual reactions that lead to the overall reaction.
Step 1: Write the given reactions in the appropriate order and manipulate them as needed to make the desired reaction.
1/2N2(g) + 1/2O2(g) → NO(g) ΔH = 90.3 kJ (Multiply the reaction by 2 to get the desired number of NO)
2NO(g) + Cl2(g) → 2NOCl(g) ΔH = -38.6 kJ (This is the desired reaction)
Step 2: Multiply the equations by appropriate coefficients to match the desired reaction. Also, we can flip the first equation to get CO(g) on the product side.
2 * (1/2N2(g) + 1/2O2(g) → NO(g)) ΔH = 2 * 90.3 kJ
2NO(g) + Cl2(g) → 2NOCl(g) ΔH = -38.6 kJ
N2(g) + O2(g) → 2NO(g) ΔH = 2 * 90.3 kJ
Step 3: Add the manipulated reactions together, canceling out any intermediate species (NO(g) in this case) that appear on both sides of the equation.
N2(g) + O2(g) + Cl2(g) → N2(g) + O2(g) + Cl(g) ΔH = 2 * 90.3 kJ + (-38.6 kJ)
Simplifying the equation and applying the values of the given reactions:
N2(g) + O2(g) + Cl2(g) → N2(g) + O2(g) + Cl(g) ΔH_rxn = 2 * 90.3 kJ + (-38.6 kJ) = 142 kJ.
Therefore, the ΔH_rxn for the given reaction is 142 kJ.