Given:
Equation #1: 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O; ΔH of reaction = -1170 kJ
Equation #2: 4NH3(g) + 3O2(g) = 2N2(g) + 6H2O; ΔH of reaction = -1530 kJ
Then what is the ΔH of reaction for: N2(g) + O2(g) = 2NO(g)? Enter a numerical value below and be sure to include a minus sign if needed. The error interval is +/- 5 kJ. The units are kJ.
To find the ΔH of reaction for the given equation N2(g) + O2(g) = 2NO(g), we can use the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the pathway between the initial and final states, as long as the initial and final conditions are the same.
To calculate the ΔH of the given equation, we need to manipulate the given equations so that adding them together will give us the desired equation.
1. Reverse Equation #1: 4NO(g) + 6H2O = 4NH3(g) + 5O2(g); ΔH = +1170 kJ (the sign of ΔH changes when the equation is reversed)
2. Multiply Equation #2 by 2: 8NH3(g) + 6O2(g) = 4N2(g) + 12H2O; ΔH = -3060 kJ (multiplied by 2)
Now, let's add the two manipulated equations together to obtain the desired equation:
4NO(g) + 6H2O + 8NH3(g) + 6O2(g) = 4NH3(g) + 5O2(g) + 4N2(g) + 12H2O
Simplifying the equation gives:
4NO(g) + 8NH3(g) + 6O2(g) = 4N2(g) + 17H2O + 5O2(g)
Note that the 17H2O on the right side of the equation cancels with the 6H2O on the left side, and the 8NH3 on the left side cancels with 4NH3 on the right side. This leaves us with:
4NO(g) + 6O2(g) = 4N2(g) + 5O2(g)
The ΔH for this equation is the sum of the ΔH values from the manipulated equations:
ΔH = (+1170 kJ) + (-3060 kJ) = -1890 kJ
Therefore, the ΔH of reaction for N2(g) + O2(g) = 2NO(g) is -1890 kJ.
To find the ΔH of reaction for the equation N2(g) + O2(g) = 2NO(g), we can use Hess's Law. Hess's Law states that the enthalpy change of a reaction is the same regardless of the pathway taken.
In this case, we can use the given equations #1 and #2 to manipulate them and obtain the desired equation.
1. Equation #1 multiplied by 2: 8NH3(g) + 10O2(g) = 8NO(g) + 12H2O; ΔH of reaction = -2340 kJ.
2. Equation #2 multiplied by 2: 8NH3(g) + 6O2(g) = 4N2(g) + 12H2O; ΔH of reaction = -3060 kJ.
Now, we add these two equations together to cancel out the common species on both sides of the equations.
8NH3(g) + 10O2(g) + 8NH3(g) + 6O2(g) = 8NO(g) + 12H2O + 4N2(g) + 12H2O.
Simplifying the equation gives:
16NH3(g) + 16O2(g) = 8NO(g) + 4N2(g) + 24H2O.
Applying stoichiometry, we can see that:
2(N2(g) + O2(g)) = 8NO(g) + 4N2(g) + 24H2O.
Now, we can see that 2NO(g) is a product on the right side of the equation, so we can remove it:
2(N2(g) + O2(g)) - 8NO(g) = 4N2(g) + 24H2O.
Now, rearrange the equation for the desired equation:
N2(g) + O2(g) = 4N2(g) + 24H2O - 8NO(g).
Finally, we can calculate the ΔH of reaction for the desired equation:
ΔH of reaction = (ΔH of equation 1 + ΔH of equation 2 - ΔH of equation 3)
ΔH of reaction = (-2340 kJ + (-3060 kJ) - (-1170 kJ))
ΔH of reaction = -2340 kJ - 3060 kJ + 1170 kJ.
Calculating this gives us the ΔH of reaction for the desired equation:
ΔH of reaction = -4230 kJ + 1170 kJ
ΔH of reaction = -3060 kJ.
Therefore, the ΔH of reaction for N2(g) + O2(g) = 2NO(g) is -3060 kJ.
Add equation 1 to the reverse of equation 2, then divide the final equation by 2.
When reversing an equation change the sign of dH. When dividing an equation by 2 divide dH by 2. When adding equations add dH values for each.