# Calculating Fraction of A Dissociated at 1000K with 1 atm and 10^-3 atm Pressure

• cantgetno
In summary, the conversation discusses the calculation of the fraction of A dissociated at 1000K with an equilibrium constant of 1x10^-2 and total pressures of 1 atm and 10^-3 atm. The equations PV=nRT and B^2/A=Ke are mentioned, and there is confusion about solving in terms of partial pressures. The final solution involves using a simultaneous equation and solving for pA and pB, resulting in pA=0.905atm and pB=0.095atm.
cantgetno

## Homework Statement

The Equilibrium constant for A(g) -> 2B(g)
is 1x10^-2 at 1000k. Calculate the fraction on A which is dissociated at 1000k if the total pressure is
1 atm
10^-3 atm

## Homework Equations

PV=nRT
B^2 / A = Ke (1x10^-2)

## The Attempt at a Solution

for A PV=1x8.714x1000 = 8.714x10^3
for B PV=2x8.714x1000 = 17.428

All i can work out from this is that there's gunna be a 1:2 ratio? I am completely lost and my minds gone blank. Thanks in advance for the help

Solve in terms of partial pressures. What is sum of pA and pB?

surely it would be pA+2pB=1atm ?
and I am not very sure how to solve in terms of partial pressures

would i have
kp= pB^2 / pA
so 0.01 = pB^2 / pA

I've tried using a simultaneous equation with pA+2pB=1
and just end up with rubbish

Why pA + 2 pB? Do you know definition of partial pressure?

because there's 2 moles of it? i thought that was right, my bad.

pA+pB=1atm that right?
would i be right to use a simlt equ then?

2*yes.

ok using the equations i end up with the quadratic:
pB^2 + 0.01pB-0.01=0 so pB=0.095atm
so pA=0.905atm
and i can work out for the other atmthanks

## 1. How do you calculate the fraction of a substance that is dissociated at 1000K?

The fraction of a substance that is dissociated at 1000K can be calculated using the equilibrium constant expression, where the dissociation constant (K) is equal to the product of the concentration of the dissociated particles over the concentration of the undissociated particles. This can also be written as [Dissociated]/[Undissociated].

## 2. What is the significance of 1000K in this calculation?

1000K is the temperature at which the calculation is being performed. The fraction of dissociated particles can vary depending on the temperature, so it is important to specify the temperature at which the calculation is being done.

## 3. How does pressure affect the fraction of dissociated particles at 1000K?

The pressure has a direct impact on the fraction of dissociated particles at 1000K because it affects the concentration of the particles. According to Le Chatelier's principle, an increase in pressure will shift the equilibrium towards the side with fewer moles of gas, resulting in a decrease in the fraction of dissociated particles.

## 4. Why is it important to specify the pressure in this calculation?

The pressure is an important factor in this calculation because it affects the concentration of the particles, which in turn affects the equilibrium constant and the fraction of dissociated particles. Without specifying the pressure, the calculation will not be accurate.

## 5. Can this calculation be used for any substance at 1000K with 1 atm and 10^-3 atm pressure?

No, this calculation can only be used for substances that undergo dissociation at 1000K and follow the ideal gas law. Additionally, the pressure must be specified in order to accurately calculate the fraction of dissociated particles.

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