Question

C2H4+H2<-->C2H6
Kc=9.8*10^-3
Initial concentration of C2H4: Reaction a: 0.1 M
Initial concentration of H2: Reaction a: 0.2 M
Initial concentration of C2H4: Reaction b:0.5 M
Initial concentration of H2: Reaction b: 0.05 M

What are the equilibrium concentrations of C2H4, H2, and C2H6?

Answer 1

why do you have 2 initial concentrations of \(H_2\)?

Answer 2

It gives reaction a and reaction b for both @aaronq

Answer 3

oh i see, they're two different scenarios.

You need to make an ICE table, and plug your values from the equilibrium line into the equilibrium expression for the reaction and solve it.

Answer 4

Intial Concentration: 0.1 0.2 -
Equilbrium 0.1-x, 0.2-x, x
Kc=[C2H6]/[C2H4][H2]
9.8*10^-3=[x]/[0.1-x][0.2-x]
Kc is a small number this indicates that more reactants are than products. X is negligible compared to concentration of reactants.
[x]=0.1*0.2*9.8*10^-3
x=0.000196 mol/dm^3
you could try out reaction b

Answer 5

Ok the ones I'm struggling with are the C2H6 equilibrium concentrations @Abmon98

Answer 6

x represent the concentration of C2H6 at equilbrium