- #1

krootox217

- 51

- 2

## Homework Statement

I have the following task:

A chemical reaction 2 A(g) <-> B(g) + C(g)

has an equilibrium constant defined by: ln=0.8+ 1050/T−1,3.10

^{5}/T²

a) What is K at 400K? Calculate ΔrG

^{0}

In a volume of 5L you get 2 moles of A and 0.5 moles of B. Reaction still proceeds at 400K. Consider the gas as ideal.

b) What are the amount of compounds when equilibrium is reached? What is the partial pressure of each compounds? What is the average molar mass of the gas mixture?

c) Calculate the kinetic energy of each gaseous compound.

I already have the solutions, but I can't find the right one for b)

**K = 13.633**

at equilibrium, Molar fraction of A = 0.12 ; B = 0.54 ; C = 0.34

Therefore Partial pressure of A = 2.10^5 Pa ; B = 8,98.10^5 Pa ; C = 5,65.10^5 Pa

average molar mass of mixture : 39.2 g/mol

Ek A = 1,497 kJ ; B = 6,734 kJ ; C = 4,24 kJ

at equilibrium, Molar fraction of A = 0.12 ; B = 0.54 ; C = 0.34

Therefore Partial pressure of A = 2.10^5 Pa ; B = 8,98.10^5 Pa ; C = 5,65.10^5 Pa

average molar mass of mixture : 39.2 g/mol

Ek A = 1,497 kJ ; B = 6,734 kJ ; C = 4,24 kJ

I get the right value for K, and then I tried it this way:

A=2mol-2x

B=0.5mol+x

C=x

13.6=K=(product)/(starting material)=((x)*(0.5+x))/(2-2x) --> x=0.8542

and then I calculated 2mol-2*X= 0.2916 mol

0.2916mol/2.5 mol = 0.12 = molar fraction of A

and I got the right fractions for B and C, and also the right partial pressures

but how do i calculate the average molar mass of the mixture?

## Homework Equations

See above

## The Attempt at a Solution

See above