Chemistry help!

Question

Consider the reaction below:

N₂(g) + 2 O₂(g) ------- 2 NO₂(g)         ΔH = +66.4 kJ

What would the enthalpy change for the following reaction be?

4 NO₂(g)  -------  2 N₂(g) + 4 O₂(g)      ΔH = ???



a. +66.4 kJ

b. -66.4 kJ

c. +132.8 kJ

d. -132.8 kJ

e. -33.4 kJ

Answer 1

The first reaction with ΔH = +66.4 kJ indicates that this reaction is endothermic, as in the reaction takes in 66.4 kJ to form 2 moles of 2 NO2(g). This is indicates by the positive sign of the enthalpy of reaction.

Imagine ΔH is a reactant in the first reaction because the reactants require ΔH = +66.4 kJ to react. If we switch the order of the reaction so that NO2(g) breaks down to N2 and O2, the ΔH will switch to the product side, suggesting that the reaction releases 66.4 kJ of energy instead. To find this new enthalpy, we switch the sign of ΔH, so it becomes ΔH = -66.4 kJ per mole of 2 moles of NO2. This reaction is exothermic.

Finally, if we want to find the enthalpy of the exothermic reaction with 4 moles of NO2, we multiply ΔH by 2. So, the final ΔH is -132.8 kJ.

Answer 2

If

N2(g) + 2O2(g) --------> 2NO2(g) (delta)H = +66.4kJ

then

2NO2(g) -----------> N2(g) + 2O2(g) (delta)H = -66.4kJ

If this reaction occurs twice, then the change in enthalpy ((delta)H) will be double:

2[2NO2(g) -----------> N2(g) + 2O2(g)]

-> 4NO2(g) ----------> 2N2(g) +4O2(g)

(delta)H = (-66.4 kJ)*2 = -132.8 kJ

So the change in enthalpy is -132.8 kJ.