What are the Mass and Empirical Formula of the Combusted Hydrocarbon?

AI Thread Summary
The discussion revolves around calculating the mass and empirical formula of a hydrocarbon combusted in oxygen, yielding specific amounts of CO2 and H2O. The mass of the hydrocarbon is determined to be approximately 6.47 g, with the empirical formula calculated as C1H1. Additionally, the heat of combustion per empirical formula unit is found to be around 624.5 kJ. A separate reaction involving the equilibrium constant is also mentioned, with a focus on understanding the relationship between forward and backward reaction rates. The discussion highlights the need for clarity on kinetic interpretations of equilibrium in chemical reactions.
fabrave21
Messages
2
Reaction score
0
2 question I don't get ... =\ any help would be appreciated.
I don't even know where to start ...

Homework Statement



2. A sample of a hydrocarbon is combusted completely in O2 to produce 21.83 g CO2, 4.47 g H2O and 311 kJ of heat. A) What is the mass of the hydrocarbon sample that was combusted? B) What is the empirical formula of the hydrocarbon? C) Calculate the value of ΔHfº per empirical-formula unit of the hydrocarbon.7. Suppose you have some reaction: 2A ---> B and B ---> 2A The rate for the formation of B from A is 3.8 x 102 s-1 and the rate for the formation of A from B is 8.6 x 10-1 s-1. A) What is the equilibrium constant for the equilibrium reaction, 2A <----> B ?
B) Suppose that you started with 2 moles of A and at some later time you measure .9 moles of B in a total volume of 1L, is the reaction at equilibrium? Explain.

Homework Equations


The Attempt at a Solution



2. mols C = 21.83/44 =.496

mols H = 4.47/18.0 =,248 x2 =.496

mols = g/Mwt then mols x Mwt = g

The mass of C is .496 x12.0 = .5.97 g C
The mass of H = .496 x 1.00 = ,496 g H
Total mass is 6.47 g for the hydrocarbon

The empirical formula is the ratio of the mols of C and H which makes it C1H1

6.47/ 13 = .496 mols
heat of combustion per formula weight is 311KJ/ .498 mols

= 624.497992 kJ

7. ?
 
Last edited:
Physics news on Phys.org
figured out number 2. Still don't get number 7
 
Have you heard about kinetic intepretation of equilibrium?

Write expressions for
- reaction equilibrium
- forward reaction speed
- backward reaction speed

Compare denominator with forward reaction speed and denominator with backward reaction speed. Do you think you can combine rate coefficients to get equilibrium constnats?
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

Similar threads

Replies
8
Views
3K
Replies
1
Views
3K
Replies
5
Views
2K
Replies
4
Views
3K
Replies
2
Views
1K
Replies
9
Views
2K
Replies
2
Views
14K
Back
Top