1
$\begingroup$

There is a reaction: $$\ce {ClF3 <=> ClF + F2}$$ with $\ce {K_\mathrm{c} = 2.48 x 10^{-3}}$ Initially, $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, and $\ce {[F2] = 0.02 M} $. $\ce {F2}$ is added until the concentration of $\ce {F2} $ is $\ce {0.1 M} $.
What is the concentration of these substances after equilibrium?

My attempt: $\ce {ClF3 <=> ClF + F2} $

Initial concentration: $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, $\ce {[F2] = 0.1 M} $
Concentration that react: $\ce {[ClF3] = +x M} $, $\ce {[ClF] = -x M} $, $\ce {[F2] = -x M}$
Concentration after equilibrium: $\ce {[ClF3] = 0.25+x M} $, $\ce {[ClF] = 0.031-x M} $, $\ce {[F2] = 0.1-x M} $
So, $\ce {Kc = [ClF].[F2].[ClF3]^{-1}} $

I applied the concentration after equilibrium in $\ce {Kc} $, so I got:
$\ce {2.48.10^{-3} = [0.1-x].[0.031-x].[0.25+x]^{-1}} $

After calculate the equation above, I got $\ce {x = 0.1111 M}$.

Then my question is: if $\ce {x = 0.1111 M} $, so $\ce {[ClF] = -0.0801 M} $ and $\ce {[F2] = -0.0111 M} $. The concentration is negative. Is it possible? Are my steps incorrect or the answer is correct?

$\endgroup$
4
$\begingroup$

The equation $2.48\times 10^{-3} = \frac{(0.1-x)(0.031-x)}{0.25+x}$ is correct, but it simplifies to a second degree polynomial, so it has two solutions, not one. You just picked the incorrect root (negative concentrations have no physical meaning). Can you find the right answer now?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.