Kasaki

Moggi's Assistant
Jun 1, 2010
13,750
Go ahead Kasaki, how can I help you?



Aaah I got you :D

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
 
Apr 15, 2006
56,640
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
:lol: Enthalpy... fuck me, it's been ages since I heard that term.
 

Delle Alpi

Chemical Dean
May 26, 2009
8,679
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,
 

Kasaki

Moggi's Assistant
Jun 1, 2010
13,750
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

Chemical Dean
May 26, 2009
8,679
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

Moggi's Assistant
Jun 1, 2010
13,750
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

Chemical Dean
May 26, 2009
8,679
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

Moggi's Assistant
Jun 1, 2010
13,750
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
 

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