Calculate DHo for the oxidation of nitric oxide to nitrogen dioxide,

2 NO(g) + O2(g) --> 2 NO2(g),
from the following data:

DHof[NO(g)] = +90.25 kJ
DHof[NO2(g)] = +33.18 kJ

im confused...can someone help me.

i got -114.14 kj

can someone confirm this?

-114.14 kJ is correct. From the data given, you have 4 significant figures. If your prof is a stickler for s.f., you should round the number to -114.1 kJ.

But how did you get that?

Of course, I'd be happy to help! Let's start by finding the enthalpy change of formation (∆Hof) for the reaction.

To do this, we can use the equation:

∆Hof (2 NO2(g)) - ∆Hof (2 NO(g)) - ∆Hof (O2(g)) = ∆Hof (2 NO(g) + O2(g) → 2 NO2(g))

Given that ∆Hof[NO(g)] = +90.25 kJ and ∆Hof[NO2(g)] = +33.18 kJ, we can plug these values into the equation:

+2 * ∆Hof[NO2(g)] - 2 * ∆Hof[NO(g)] - ∆Hof[O2(g)] = ∆Hof[2 NO(g) + O2(g) → 2 NO2(g)]

Substituting the given values:

+2 * (+33.18 kJ) - 2 * (+90.25 kJ) - ∆Hof[O2(g)] = ∆Hof[2 NO(g) + O2(g) → 2 NO2(g)]

Simplifying:

+66.36 kJ - 180.50 kJ - ∆Hof[O2(g)] = ∆Hof[2 NO(g) + O2(g) → 2 NO2(g)]

=-114.14 kJ - ∆Hof[O2(g)] = ∆Hof[2 NO(g) + O2(g) → 2 NO2(g)]

Now, we just need to find the value for ∆Hof[O2(g)]. Since it was not given in the data, we can refer to standard tables, such as the average bond enthalpy. The bond enthalpy of O=O is approximately 498 kJ/mol.

∆Hof(O2(g)) = 498 kJ/mol

Now, we can substitute this value into our equation:

-114.14 kJ - 498 kJ/mol = ∆Hof[2 NO(g) + O2(g) → 2 NO2(g)]

Simplifying:

∆Hof[2 NO(g) + O2(g) → 2 NO2(g)] ≈ -612.14 kJ/mol

So, the enthalpy change (∆Ho) for the oxidation of nitric oxide to nitrogen dioxide is approximately -612.14 kJ.

To calculate the standard enthalpy change (ΔH°) for a reaction, you can use the following equation:

ΔH° = Σ(ΔHof(products)) - Σ(ΔHof(reactants))

Where ΔHof is the standard enthalpy of formation, which is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.

Given the data provided:

ΔHof[NO(g)] = +90.25 kJ
ΔHof[NO2(g)] = +33.18 kJ

We can now express the enthalpy change as:

ΔH° = [2 x ΔHof(NO2)] - [2 x ΔHof(NO)] - [ΔHof(O2)]

Substituting the values:

ΔH° = [2 x (+33.18 kJ)] - [2 x (+90.25 kJ)] - [0 kJ]

ΔH° = +66.36 kJ - 180.5 kJ - 0 kJ

ΔH° = -114.14 kJ

Therefore, the standard enthalpy change (ΔH°) for the reaction 2 NO(g) + O2(g) → 2 NO2(g) is -114.14 kJ.