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$\pu{2.0M}$ of $\ce{I2}$ and $\pu{2.0M}$ of $\ce{Br2}$ are initially present in a reaction mixture. Calculate the equilibrium concentration of $[\ce{I2}]$ , $[\ce{Br2}]$ and $[\ce{IBr}]$ at $\pu{425K}$. $K_c=100$

Equation: $\ce{I2 + Br2 -> 2IBr }$


  • 1st method

Equation:$\ce{I2 \ \ \ + \ \ \ Br2 ->\ \ \ \ 2IBr}$

Initial: $\ \ \pu{2.0M}\ \ \ \ \ \ \pu{2.0M} \ \ \ \ \ \ \ \ \ \ 0$

Change: $\ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ 2x$

Equilib: $\ \ 2-x \ \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ 2x$

$K_\mathrm c =\frac{(2x)^2}{(2-x)^2}=100$


  • 2nd method what if I separate $2\ce{IBr}$

Equation: $\ce{I2 \ \ \ + \ \ \ Br2 ->\ \ \ IBr \ \ + \ \ \ IBr }$

Initial: $\pu{2.0M}\ \ \ \ \ \ \ \pu{2.0M}\ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0$

Change: $\ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x$

Equilib: $2-x \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ x $

$K_\mathrm c =\frac{(x)(x)}{(2-x)^2}=100$

So which one is actually right and which one is wrong? Why?

The idea to use the second method came to my mind because of the the definition:

The equilibrium constant is a ratio of the concentration of the products to the concentration of the reactants

If $\ce{aA +bB \ —> cC + dD}$ (1) $K_\mathrm c = \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$

If I separate : $a\ce{A} = \ce{A}+\ce{A}+...+\ce{A}$ ($\ce{A} a$ times )

Same thing with $b\ce{B}, c\ce{C}$ and $d\ce{D}$, the equation (1) becomes:

$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$

$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}$=$ \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$ (Proved)

So, by using the second method, how and why is it not right ?

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    $\begingroup$ Both methods are right and both are the same, except that you are not using the 2nd one right. Just where did those little $x^2$'s come from? $\endgroup$ – Ivan Neretin Feb 23 '18 at 5:21
  • $\begingroup$ Oh it’s actually (x)(x) ! Am confused $\endgroup$ – November ft Blue Feb 23 '18 at 6:16
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    $\begingroup$ No, it's actually (2x)(2x), because the concentration of IBr is 2x. $\endgroup$ – Ivan Neretin Feb 23 '18 at 6:29
  • $\begingroup$ What I think is 2x is the concentration of 2IBr ( 2 molecules of IBr ) not just IBr because in the “ change “ section we already balance the molecule. $\endgroup$ – November ft Blue Feb 24 '18 at 4:57
  • $\begingroup$ There is no such thing as concentration of 2IBr. $\endgroup$ – Ivan Neretin Feb 24 '18 at 5:04
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So which one is actually right and which one is wrong? Why?

The second method is wrong. In your second method, you've written: $\ce{I2 \+ Br2 ->IBr + IBr }$ and then assigned the concentration $x$ and $x$ to each of the two (identical!) $\ce{IBr}$'s. Till now, it's fine.

The $K_\mathrm c$ has been used by you as:

$$K_\mathrm c=\frac{[\ce{IBr}][\ce{IBr}]}{[\ce{I2}][\ce{Br2}]}$$

which is okay but a better way to write it would have been

$$K_\mathrm c=\frac{[\ce{IBr}]^2}{[\ce{I2}][\ce{Br2}]}$$

as you've taken the single concentration term for $\ce{IBr}$ (=$2x$) from the correct formula, split it up into two parts ($x$ and $x$), and then used those two parts separately in $K_\mathrm c$, as if there were not one but two distinct $\ce{IBr}$'s being formed! It is just confusing.

Anyways, the first formula will still work. The main problem here is that $[\ce{IBr}]=\color{blue}{\text{total }}\text{concentration of}~\ce{IBr}$. When you put any reactant/product inside the square brackets, you're supposed to write their total concentration, even if they have been split up in parts. So, when in the derivation of $K_\mathrm c$, you write:

$$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$$

for the reaction: $$\ce{aA + bB —> cC + dD}$$

each of those individual $[\ce{C}]$'s will refer to the total concentration of $\ce{C}$. That means, as $\ce{C}$ was separately written $c$ times by you in the reaction, the final $[\ce{C}]$ that you use will be the sum of all those $c$ $[\ce{C}]$s.

So, in the end of your second method, while applying the correct formula for $K_\mathrm c$, you must add those $x$ and $x$ together, and then write:

$$K_\mathrm c=\frac{(x+x)\times(x+x)}{(2-x)\times(2-x)}$$

into the formula you were using.

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