Given the Data:
N2(g) + O2g) = 2NO(g) delta H = 180.7kj, 2NO(g) + O2(g) = 2NO2(g) delta H = 0113.1kj, 2N2O(g) = 1N2(g) + O2(g) delta H = -163.2kj.
Using Hess’s Law calculate delta H for the reaction:
N2O(g) + NO2(g) = 3NO(g)
To calculate the delta H for the reaction N2O(g) + NO2(g) = 3NO(g) using Hess's Law, we need to rearrange and combine the given equations so that the reactants and products in the target equation appear.
Given reactions:
1) N2(g) + O2(g) = 2NO(g) (delta H = 180.7 kJ)
2) 2NO(g) + O2(g) = 2NO2(g) (delta H = 113.1 kJ)
3) 2N2O(g) = N2(g) + O2(g) (delta H = -163.2 kJ)
To start, let's reverse equation 2 to match the orientation of equation 1:
4) 2NO2(g) = 2NO(g) + O2(g) (delta H = -113.1 kJ)
Now, add equation 3 and equation 4 to cancel out NO(g):
5) 2N2O(g) + 2NO2(g) = N2(g) + 2NO(g) + O2(g) (delta H = -163.2 kJ - 113.1 kJ = -276.3 kJ)
Lastly, multiply equation 1 by 2 and equation 5 by 3 and add them to obtain the target reaction:
6) 2(N2(g) + O2(g) = 2NO(g)) + 3(2N2O(g) + 2NO2(g)) = 2N2(g) + 4NO(g) + 3N2O(g) + 3NO2(g) + 3O2(g)
(delta H = 2 * 180.7 kJ + 3 * -276.3 kJ = -132.2 kJ)
Therefore, the delta H for the reaction N2O(g) + NO2(g) = 3NO(g) is -132.2 kJ.
To calculate delta H for the reaction N2O(g) + NO2(g) = 3NO(g), we need to use the given equations and apply Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the reactions that lead to the desired reaction.
Let's break down the desired reaction into smaller steps:
1. We need 2NO(g) in the final reaction, so we can multiply the second equation by 2 and reverse its direction:
2NO(g) + 2O2(g) --> 4NO2(g) (multiply by 2, reverse direction)
2. We need 2NO2(g) in the final reaction, so we can use the second equation as is:
2NO(g) + O2(g) --> 2NO2(g) (unchanged)
3. We need N2O(g) and NO2(g) in the final reaction, so we can use the first equation and reverse its direction:
2NO(g) --> N2(g) + O2(g) (reverse direction)
Now we can sum up these equations:
2NO(g) + 2O2(g) --> 4NO2(g) (multiply by 2, reverse direction)
2NO(g) + O2(g) --> 2NO2(g)
2NO(g) --> N2(g) + O2(g) (reverse direction)
Summing these equations and canceling out common species gives us the desired reaction:
N2O(g) + NO2(g) = 3NO(g)
Now, we can calculate the overall delta H for the desired reaction by summing the delta H values of the individual equations:
[2NO(g) + 2O2(g) --> 4NO2(g)] * 2 --> delta H1
[2NO(g) + O2(g) --> 2NO2(g)] --> delta H2
[2NO(g) --> N2(g) + O2(g)] * -1 --> delta H3
delta H for the desired reaction = delta H1 + delta H2 + delta H3
Let's substitute the given delta H values:
delta H1 = (2 * 113.1) kJ = 226.2 kJ (since it was multiplied by 2)
delta H2 = 113.1 kJ
delta H3 = (-1 * -163.2) kJ = 163.2 kJ (since it was reversed)
delta H for the desired reaction = 226.2 kJ + 113.1 kJ + 163.2 kJ
= 502.5 kJ
Therefore, the delta H for the reaction N2O(g) + NO2(g) = 3NO(g) is 502.5 kJ.
multiply equation 1 by 2 and dH x 2.
Reverse equation 2 and change sign dH.
Balance equation 3(it isn't balanced now).
Add the three equations and add dHs. This will get
2NO2 + 2N2O ==> 6NO which is twice what you want. So divide the final equation by 2 and divide sum dH by 2.