Calculating Mole Fraction and Kp in Chemical Equilibrium

In summary, the conversation discusses finding the mole fraction of O2 and the value of Kp for a chemical equilibrium problem involving the reaction 2SO2(g) + O2(g) ↔ 2 SO3(g). The final answer for Kp is incorrect and needs to be converted to Kc. The best approach is to calculate the total molarity of the gas from the given pressure and subtract it from the known total S molarity to find the molarity of O2.
  • #1
rock23
3
0
Chemical equilibrium help!

For a reaction 2SO2(g) + O2(g) ↔ 2 SO3(g), 0.1mol of each SO2 and SO3 are mixed in a 2.0L flask at 27 degrees Celsius. After Equilibrium total pressure is 2.78atm.
Calculate a) The mole fraction of O2 at equilibrium
b) The value of Kp

I don't know how to find out the mole fraction of O2...
I did the problem finding Kp.. but I got a large number of 19.31...
And my Nt at equilibrium= n(4-alpha)

Any ideas??
 
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  • #2


Here is my work:

2SO2(g) + O2(g) ↔ 2 SO3(g)
2n 0 ↔ 2n
-2n@ -n@ ↔ +2n@

2n-2n@ -n@ ↔ 2n +2n@ Where Nt= 2n-2n@-n@+2n+2n@
2n(1-@) -n@ ↔ 2n(1+@) Nt=4n-n@ Nt= n(4-@)

2n(1-@)/n(4-@) -n@/(n4-@) ↔ 2n(1+@)/ (n (4-@))

2(1-@)/(4-@) Pt 1/4 Pt ↔ 2(1+@)/ (4-@))

Where Kp= (P SO3(g))^2 / ((P SO2)^2 * (P O2))

Kp= ((4(1+@)^2) / ((1-@)^2 ) ) 1/Pt

Where Nt at equilibrium= ( 2.78atm*2L) /( 0.08206 atm dm3 mol-1 k-1)(301.15)
Nt= .2249 moles

solving for @
.2449= .1(4-@)
@= 1.751

Plugging in @ to find Kp... I got 19.31... but i'ts wrong it's supposed to be 0.356
 
  • #3


try to convert the Kp to Kc.
 
  • #4


You probably haven't received help because your answer is too hard to read.
I don't know what @ or Nt mean.
And just above the input box there are symbols X2 and X2 which allow you to easily write things like [SO3]2 which help legibility.

The way I would find it easiest:

Total S (sulphur) is 0.1 M if I am not mistaken.
From the pressure work out the total molarity of the gas.
The difference is the molarity of O2.
 
Last edited:

1. What is chemical equilibrium?

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the forward and reverse reactions are happening at the same rate, resulting in no net change in the amount of reactants and products.

2. How is chemical equilibrium different from a chemical reaction?

A chemical reaction is a process where reactants are transformed into products, while chemical equilibrium is a state where the concentrations of reactants and products remain constant. In a chemical reaction, the reactions are still occurring, but in chemical equilibrium, the forward and reverse reactions are happening at the same rate, resulting in no net change in the concentrations of reactants and products.

3. What factors can affect chemical equilibrium?

The factors that can affect chemical equilibrium include temperature, pressure, and the concentrations of reactants and products. Changing any of these factors can shift the equilibrium position, resulting in a change in the concentrations of reactants and products.

4. How can you determine if a reaction is at chemical equilibrium?

There are a few ways to determine if a reaction is at chemical equilibrium. One way is to observe if the concentrations of reactants and products remain constant over time. Another way is to calculate the equilibrium constant, which can help determine the position of equilibrium and the relative concentrations of reactants and products.

5. How can you manipulate chemical equilibrium to favor the formation of a certain product?

Chemical equilibrium can be manipulated by changing the conditions of the reaction. For example, increasing the concentration of a reactant can shift the equilibrium towards the formation of products. Changing the temperature or pressure can also affect the position of equilibrium. Additionally, using a catalyst can speed up the rate of the reaction, which can also shift the equilibrium in favor of the desired product.

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