What Are You Doing Right Now? (861 Viewers)

[Kasaki]

Go ahead Kasaki, how can I help you?



Aaah I got you

Use the following reaction.
4NO2(g) + 6H2O(g) = 4NH3(g) + 7O2(g)

a. Calculate the standard enthalpy change for the reaction .

This is the Given

H2O(l) enthalpy final = -285.8
H2O (g) enthalphy final = -241.82
NH3 (g) enthalpy final = -46.11
NO2 (g) enthalpy final = 33.18

I got 1133.36kJ

But I know enthalpy change is Hfinal - Hinitial

 

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[Sheik Yerbouti]

Use the following reaction.
4NO2(g) + 6H2O(g) = 4NH3(g) + 7O2(g)

a. Calculate the standard enthalpy change for the reaction .

This is the Given

H2O(l) enthalpy final = -285.8
H2O (g) enthalphy final = -241.82
NH3 (g) enthalpy final = -46.11
NO2 (g) enthalpy final = 33.18

I got 1133.36kJ

But I know enthalpy change is Hfinal - Hinitial

Enthalpy... fuck me, it's been ages since I heard that term.

 

[Kasaki]

Enthalpy... fuck me, it's been ages since I heard that term.

Care to trade places? This class stole my virginity

 

[Sheik Yerbouti]

Naah. I was pretty bad at chemistry.

 

[X Æ A-12]

HOW MANY POTS HAVE YOU SMOKIN'?

Nigga, I'm moked.

 

[Delle Alpi]

Use the following reaction.
4NO2(g) + 6H2O(g) = 4NH3(g) + 7O2(g)

a. Calculate the standard enthalpy change for the reaction .

This is the Given

H2O(l) enthalpy final = -285.8
H2O (g) enthalphy final = -241.82
NH3 (g) enthalpy final = -46.11
NO2 (g) enthalpy final = 33.18

I got 1133.36kJ

But I know enthalpy change is Hfinal - Hinitial

The standard enthalpy = the total enthalpy of the products after multiplying each by it's stoichemetric coefficient - the total enthalpy of the reactants after multiplying each by it's stoichemetric coefficient

stoichemetric coefficient = the number of moles of each compound,

 

[Delle Alpi]

Kasaki, are you still around? Did it make sense?

 

[Kasaki]

The standard enthalpy = the total enthalpy of the products after multiplying each by it's stoichemetric coefficient - the total enthalpy of the reactants after multiplying each by it's stoichemetric coefficient

stoichemetric coefficient = the number of moles of each compound,

thanks , makes a lil more sense now, guess Ill take this studying day by day, can't expect to fully understand it all in one day right?

 

[Delle Alpi]

thanks , makes a lil more sense now, guess Ill take this studying day by day, can't expect to fully understand it all in one day right?

You are welcome. So each one of the reactant and the products has an enthalpy for example

5X + 3Y ------> 2A + 3C

X = 123
Y = 232
A = 134
C = 154
Then the standard enthalpy of the reaction should be (2x 134) + (3x154) - (5x123)+(3x232) = answer.

P.S Always make sure the reaction is balanced before calculating anything

 

[Kasaki]

You are welcome. So each one of the reactant and the products has an enthalpy for example

5X + 3Y ------> 2A + 3C

X = 123
Y = 232
A = 134
C = 154
Then the standard enthalpy of the reaction should be (2x 134) + (3x154) - (5x123)+(3x232) = answer.

P.S Always make sure the reaction is balanced before calculating anything

Putting it in letters helped alot more!

In regards to balancing equation when does factoring and dividing by 1/2 come into play

 

[Delle Alpi]

Putting it in letters helped alot more!

In regards to balancing equation when does factoring and dividing by 1/2 come into play

in balancing equation, we don't like to have 1/2s, it should be always an integer. Make sure you have the exact number of atoms on each side, and only integers. Once you balance the equation, then you get the total enthalpy for the products after multiplying each by the stoichiometric coefficients and then subtract the total enthalpy of the reactants

 

[Kasaki]

in balancing equation, we don't like to have 1/2s, it should be always an integer. Make sure you have the exact number of atoms on each side, and only integers. Once you balance the equation, then you get the total enthalpy for the products after multiplying each by the stoichiometric coefficients and then subtract the total enthalpy of the reactants

coolio , thanks a lot. Where were u the other day when I was taking my online chem quiz lol

 

[Delle Alpi]

coolio , thanks a lot. Where were u the other day when I was taking my online chem quiz lol

I was probably in the lab!
Good luck with your exam and just keep working on solving problem sets, you should be alright.

 

[Christina]

http://twitter.com/#!/CherylKerl

 

[X Æ A-12]

I don't get it.

 

[Christina]

Have you ever heard Cheryl Cole speak?

 

[X Æ A-12]

I have not.

 

[Kasaki]

Have you ever heard Cheryl Cole speak?

no

 

[Kate]

Have you ever heard Cheryl Cole speak?

I got it, and

 

[Sheik Yerbouti]

Have you ever heard Cheryl Cole speak?

Why would anyone want to hear a woman\girl like that speak? The most I would permit is intense moaning and "Oh God".