Using Hess law, calculate ΔH for the equation: O3 (g) +2NO2 (g)→ N2O5 (g) + O2 (g)
Given the following equations:
N2O5 (g) rightwards arrowNO3 (g) + NO2 (g) ΔH=+ 44.3 kJ
NO3 (g) + O2 (g) rightwards arrowO3 (g) + NO2 (g) ΔH=-20.1kJ
To use Hess's law, we need to manipulate the given equations so that they add up to the desired equation:
N2O5 (g) rightwards arrowNO3 (g) + NO2 (g) (multiply by 1)
2NO3 (g) + O2 (g) rightwards arrow2O3 (g) + 2NO2 (g) (multiply by 2 and flip the direction)
O3 (g) + 2NO2 (g) rightwards arrowN2O5 (g) + O2 (g) (add the two equations)
Now we can add the enthalpies of the individual reactions to get the total enthalpy change for the desired reaction:
ΔH = ΔH1 + ΔH2
ΔH1 = +44.3 kJ (from the first equation)
ΔH2 = -20.1 kJ (from the second equation, multiplied by 2 and flipped)
ΔH = +44.3 kJ - 2(-20.1 kJ) = +84.5 kJ
Therefore, the enthalpy change for the reaction O3 (g) + 2NO2 (g) → N2O5 (g) + O2 (g) is +84.5 kJ.
To calculate the ΔH for the given equation using Hess's Law, we can use the following steps:
Step 1: Reverse the first equation and its ΔH value:
NO3 (g) + NO2 (g) rightwards arrow N2O5 (g) ΔH=-44.3 kJ
Step 2: Multiply the second equation by 2 to balance the number of moles of NO2:
2(NO3 (g) + O2 (g) rightwards arrow O3 (g) + 2NO2 (g)) ΔH=-2(-20.1 kJ) = +40.2 kJ
Step 3: Multiply the first equation by 2 to match the number of moles of NO2 in the target equation:
2(N2O5 (g) rightwards arrow NO3 (g) + NO2 (g)) ΔH=2(44.3 kJ) = +88.6 kJ
Step 4: Add all the equations together to obtain the overall equation:
2(NO3 (g) + O2 (g) rightwards arrow O3 (g) + 2NO2 (g)) + 2(N2O5 (g) rightwards arrow NO3 (g) + NO2 (g)) = O3 (g) + 2NO2 (g) rightwards arrow N2O5 (g) + O2 (g)
ΔH = +40.2 kJ + 88.6 kJ = 128.8 kJ
Therefore, the ΔH for the given equation is +128.8 kJ.