Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles of oxygen gas and 0.625 moles of sulfur trioxide gas are present in a 1.50-liter reaction vessel at equilibrium. 2SO2(g) + O2 (g) 2SO3 (g)

Question

Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles of oxygen gas and 0.625 moles of sulfur trioxide gas are present in a 1.50-liter reaction vessel at equilibrium. 2SO2(g) + O2 (g)  2SO3 (g)

aaronq · Answer

all you need to do is write the Kc equilibrium expression and plug your values in. (then obviously evaluate it)

anonymous · Answer

but how do you write the Kc expression? Can you show me?

aaronq · Answer

so if you have a reaction:

aA + bB <-> cC + dD

where a,b,c,d (lower case) are the stoichiometric coefficients and A,B,C,D are the species, then:

$$K _{c}=\frac{ [C]^{c}[D]^{d}  }{ [A]^{a} [B]^{b} }$$

aaronq · Answer

this should technically be Kp, since they're gases, but it's the same thing, essentially.

anonymous · Answer

so it would look something like:    = {SO3}^2 / {SO2}^2 {O2}?

aaronq · Answer

yeah, exactly. now find the concentration of each gas in the vessel.

concentration=moles/V

anonymous · Answer

so when I find the concentrations, do I plug them into the Kc equation?

aaronq · Answer

yep

anonymous · Answer

I get .0461, but the available answers are- 
		
Kc = 3.91 x 10-1
		
Kc = 4.34 x 101
		
Kc = 6.47 x 101
		
Kc = 1.47 x 102

What do you say?

anonymous · Answer

btw, all those are in scientific notation, to the numbers after 10 are exponents

aaronq · Answer

i got 65 as an answer.. which would be Kc = 6.47 x 10^1

anonymous · Answer

oh woops... I put the order backwards in my calculator. Thanks for all the help!

aaronq · Answer

haha gotta watch those computations ! no problem dude