Heat of Formation Q

[ModusProbandi]

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2 H2(g) + 2 NO(g)

2 H2O(g) + N2(g)​

Find NO's heat of formation.
DeltaH = –664 kJ/mol for the above reaction.
heat of formation of H2O(g) is –242 kJ/mol

This is the one legit question I got on a TPR test due to content loss. Halp!
I chose 422. It was wrong. I'm probably gonna be embarrassed by asking this question.

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[Mcat35]

2 H2(g) + 2 NO(g)

2 H2O(g) + N2(g)​

Find NO's heat of formation.
DeltaH = –664 kJ/mol for the above reaction.
heat of formation of H2O(g) is –242 kJ/mol

This is the one legit question I got on a TPR test due to content loss. Halp!
I chose 422. It was wrong. I'm probably gonna be embarrassed by asking this question.

​

Heat of formation reaction = Heat of formation of products - Heat of formation of reactants. Remember when you would multiply heat of formation of products and reactants by their stoichiometric coefficients.

so you would end up with -664 = 2(-242) - 2 (heat of formation of NO)

-664 + 484 = -180 kj/mol
this is heat of formation for 2x moles.

for one mole NO you would do - 180kj/mol / - 2 (NO) = 90 kj/mol

Also remember heat of formation of element in standard state is 0. So you don't have to do anything to the N2, or H2 since they are in their standard state.

 

[ModusProbandi]

Heat of formation reaction = Heat of formation of products - Heat of formation of reactants. Remember when you would multiply heat of formation of products and reactants by their stoichiometric coefficients.

so you would end up with -664 = 2(-242) - 2 (heat of formation of NO)

-664 + 484 = 180 kj/mol
this is heat of formation for 2x moles.

for one mole NO you would do 180kj/mol / 2 = 90 kj/mol

Also remember heat of formation of element in standard state is 0. So you don't have to do anything to the N2, or H2 since they are in their standard state.

But why isn't it -664 x 2, because aren't we are talking about 2 moles in such a particular reaction, the enthalpy change per mole is constant but not the enthalpy change itself?

 

[Mcat35]

But why isn't it -664 x 2, because aren't we are talking about 2 moles in such a particular reaction, the enthalpy change per mole is constant but not the enthalpy change itself?

when they say Delta H for the reaction is based on the balanced equation. This would include total heat of formation of products - reactants.

 

[ModusProbandi]

when they say delta h for the reaction is based on the balanced equation. This would include total heat of formation of products - reactants.

Nevermind, I got it. /\H/mol for a reaction is technically the same as saying /\H for 2 mols of NO, /\H for 2 moles of H2O, etc and that's where I went wrong. /\H/mol is the same as /\H when looking at the lowest whole number ratios for the reactants/products.

 

[ilovemcat]

Sorry, I know you figured this out. Was just wondering though if the answer was -90kJ/mol.

 

[ModusProbandi]

sorry, i know you figured this out. Was just wondering though if the answer was -90kj/mol.

+90.

 

[ridethecliche]

+90.

How?

I think Mcat35 made a math error. -600+400 is - not +

Wait, nevermind. The negatives cancel out!

 

[Mcat35]

How?

I think Mcat35 made a math error. -600+400 is - not +

Wait, nevermind. The negatives cancel out!

fixed it.