2
$\begingroup$

In the formation of 3-bromo-1-butene and 1-bromo-1-butene, from $\ce{HBr}$ and $\ce{C_4H_6}$, in a gas phase:

$\ce{HBr + C_4H_6 \to 1BrC_4H_7}$

$\ce{HBr + C_4H_6 \to 3BrC_4H_7}$

the number of degrees of freedom is given by: $F=C-P+2$

I'm trying to rationalise this but finding it quite confusing. For example, can we state that there is only 1 phase, i.e. a gas phase?

And the components, is the isomer of one independent of another? I'm taking there to be 3 individual components.

My answer is then 4 degrees of freedom, which seems okay, given $p$, $t$, $X$, $V$ could vary?

I'd appreciate your help in this, thanks.

$\endgroup$
3
$\begingroup$

Let's do it step by step.

  1. $C = 2$. The reason is that you have two components: $\ce{HBr}$ and $\ce{C4H6}$. The third compound, $\ce{BrC4H7}$, is simply a combination of both of them. Remember what is $C$: It's the minimal number of components needed to describe all phases in the system. Since $\ce{BrC4H7}$ can de described as $1\times\ce{HBr} + 1\times\ce{C4H6}$, it is not an independent component.
  2. $P=1$ or $P=2$. It depends on how much the reactions has progressed. Basically, you have one phase on each side of the reaction. On the left, you have one gas phase that has a mixture of both $\ce{HBr}$ and $\ce{C4H6}$. On the right, you one one liquid (?) phase: $\ce{BrC4H7}$. So here are the possibilities:
    1. The reaction has not started yet. All you have in the one gas phase. In that case, $P=1$ and $F=2-1+2=3$.
    2. The reaction is going as written, but not completed yet. In this case not all gas is consumed yet, but liquid exists. Then, $P=2$ and $F=2-2+2=2$.
    3. The reaction is finished. All gas is now liquid. Then we're back to $P=1$ and $F=2-1+2=3$.

So what happened in stage 2? Why did you lose a degree of freedom? Well, it depends on what what the driving force for the reaction. For example: Why did gaseous $\ce{HBr}$ and $\ce{C4H6}$ combine to form liquid $\ce{BrC4H7}$? Was it because of cooling or heating? In this case, the lost degree of freedom is $t$. Was it pressure? Then the lost degree of freedom is $P$.

$X$ is a bit problematic. There is not enough information given in the question. In my opinion, $X$ was fixed to begin with: I assume that once the reaction is complete, none of the reactants remain. Thus their ratio is fixed to 1:1 and $X$ is constrained. Had it not been the case, you would also get differing $P$ in stage three because you would have two phases, not one.

$\endgroup$
  • $\begingroup$ Thank you Michael. So, if I was to keep the ratio of reactants fixed at the beginning, would cause a loss of a degree of freedom? Or would that make any difference? $\endgroup$ – Edward Dec 22 '14 at 0:17
  • $\begingroup$ @Maximilion I edited my question to address $X$ $\endgroup$ – Gimelist Dec 22 '14 at 6:38

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.