# Given the following reactions, what is the H for this reaction?

Given the following reactions, what is the H for this reaction?

NO (g) + O (g) —> NO2 (g)

NO (g) + O3 (g) —-> NO2 (g) + O2 (g) ΔH₁=199 kJ
O3 (g) —-> 3/2O2 (g) . . . . . . . . . . . . ΔH₂=142 kJ
O2 (g) —-> 2O (g) . . . . . . . . . . . . . ΔH₃=+495 kJ

To determine the enthalpy changes, Hess’ law can be applied
Hess’s Law states that if a reaction occurs in two or more steps, the enthalpy for the reaction is the sum of the enthalpies of individual steps.

The simplest steps is by using the algebraic method:
The thermochemical equation for the reaction whose ΔH is to be calculated:
NO (g) + O (g) —> NO2 (g) . . . . ΔH = x kJ/mol

Rearrange the thermochemical equation with the ΔH given,
NO (g) + O3 (g) —-> NO2 (g) + O2 (g) ΔH₁=199 kJ (this remains the same, because the stoichiometric coefficient for NO in the reactant is 1, so there’s no change)

O3 (g) —-> 3/2O2 (g) . . . . . . . . . . . . ΔH₂= 142 kJ
Inverse the reaction to cancel O3 and O2. ΔH becomes negative.
3/2O2 (g) —-> O3 (g). . . . . . . . . . . . ΔH₂= -142 kJ

O2 (g) —-> 2O (g) . . . . . . . . . . . . . ΔH₃=+495 kJ
The O from the main equation is the reactantso it is an inverse reaction, and the ΔH is negative. , The stoichiometric coefficient for O is 1, so the chemical equation needs to be divided by 2 (including ΔH)
O (g) —-> 1/2O2 (g) . . . . . . . . . . . . . ΔH₃=-247.5 kJ

You have the rearranged equations:
NO (g) + O (g) —> NO2 (g) . . . . . . . . ΔH = x kJ/mol
NO (g) + O3 (g) —-> NO2 (g) + O2 (g) ΔH₁=199 kJ
3/2O2 (g) —-> O3 (g). . . . . . . . . . . . . ΔH₂= -142 kJ
O (g) —-> 1/2O2 (g) . . . . . . . . . . . . .ΔH₃=-247.5 kJ

ΔH = ΔH₁ + ΔH₂ + ΔH₃ = 199+(-142)+(-247.5) = -190.5kJ/mol
negative value shows that heat is lost to the surroundings.

Hess’ Law problems are really fun to do. You just play around with the equations and enthalpies until you get what you want. Just remember to change the enthalpy the same way you manipulate the equations.

From the final product, you can see that you want NO and O on one side and NO2 on the other. The first equation has the NO and NO2 on the right side so you take it as is. But you want to get rid of the O3 and O2. Take the second equation and flip it. Since O3 is on both sides, it cancels out. You also have to change the sign of the enthalpy since you flipped the equation. The rest of the process is below:

NO + O = NO2

NO + O3 = NO2 + O2 H=199kJ
3/2O2 = O3 H= -142kJ (negative)
3O = 3/2 O2 H= -742.5kJ (negative and multiplied by 3/2)
O2 = 2O H= 495kJ

Stuff cancels out and you’re left with the final equation. Then just add all the individual enthalpies and you get the enthalpy for the final reaction. In this case, its -190.5kJ

P4(g) + 10 Cl2(g) ? 4PCl5(s) ?H°rxn = ? Is a mix of the two reactions. some are flipped: Cl2 + PCl3 => PCl5 -157 kJ P4 + 6Cl2 => 4PCl3 -1207 kJ Now we prefer 10 Cl2 entire and would desire to cancel out the PCl3 with 4. So multiply the reaction by 4. 4Cl2 + 4PCl3 => 4PCl5 = -628 kJ 6 Cl2 + P4 => 4PCl3 = -1207 kJ The PCl3 cancels once you upload the reaction and the 4 an 6 Cl2 grow to be 10 Cl2. very final answer: A) -1835 kJ

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