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.


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.

  • $\begingroup$ 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

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