O3 + NO --> O2 + NO2 (all in gas state)

Calculate the change in enthalpy for the reaction at room temp. using the following data

^Hf:

O3 = 143
NO = 90
NO2 = 33

So, I have 143+90--> X + 33. I don't know what the enthalpy of O2 is. I assume you simply subtract the enthalpy of products from the enthalpy of reactants. Is that correct?

Yes, that is right.

delta H products - delta H reactants = delta H reaction. Delta H formatin for O2 is zero. However, let me point out that in your 143+90 ==> x + 33, you have reversed that process. You should have
DHrxn = 33 -(143 + 90)

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Yes, you are correct! To calculate the change in enthalpy (ΔH) for the reaction at room temperature using the given data, you would subtract the enthalpy of the reactants from the enthalpy of the products.

In this case, the balanced equation is:

O3 + NO → O2 + NO2

Given the enthalpy values:

ΔHf:
O3 = 143 kJ/mol
NO = 90 kJ/mol
NO2 = 33 kJ/mol

To find the enthalpy change (ΔH), you would subtract the sum of the enthalpies of the reactants from the sum of the enthalpies of the products. Let's calculate it step by step:

Sum of the enthalpies of the reactants:
143 kJ/mol (O3) + 90 kJ/mol (NO) = 233 kJ/mol

Sum of the enthalpies of the products:
X (O2) + 33 kJ/mol (NO2)

Now, to find the enthalpy change (ΔH), you can rewrite the balanced equation with coefficients:

1 O3 + 1 NO → 1 O2 + 1 NO2

By comparing the coefficients in the balanced equation with the stoichiometric ratios (1:1), you can see that the enthalpy change (ΔH) will be the same as the value of the enthalpy of the products:

ΔH = X (O2) + 33 kJ/mol (NO2)

Therefore, the change in enthalpy for the reaction at room temperature is X + 33 kJ/mol. However, you need to determine the value of the enthalpy of O2 in order to calculate the exact change in enthalpy (ΔH) for the reaction.