[earthnation112]

I just want to confirm if I have got the answers right to some revision questions for chemistry. Would be very grateful if someone can give productive input to what I post. Thank You

The question is:
The reaction of isobutyl bromide with sodium ethoxide is first order with respect to both reagents. When the initial concentrations of both reagents in the reaction mixture was 0.0242 M, the concentration of the sodium ethoxide fell to 0.0164 M after 60 minutes. Determine the rate constant for the reaction. The reaction was then carried out again using different initial concentrations, namely 0.0300 M of isobutyl bromide and 0.0200 M of sodium ethoxide. What was the concentration of sodium ethoxide after 60 minutes?

To work out the rate constant I used the following formula:


\begin{equation} \frac{1}{[A]} = kt + \frac{1}{[A]_0} \end{equation}
This was then rearranged to give the following:
\begin{equation} k= \frac{\frac{1}{[A]} - \frac{1}{[A]_0}}{t} \end{equation}
\begin{equation} k= \frac{\frac{1}{[0.0164]} - \frac{1}{[0.0242]_0}}{60} \end{equation}
\begin{equation} k = \frac{19.7}{60} \end{equation}
k = 0.333

Then to answer the question asking for the concentration of sodium ethoxide after 60 minutes I used the same formula that was introduced at the start:

\begin{equation} \frac{1}{[A]} = kt + \frac{1}{[A]_0} \end{equation}
\begin{equation} \frac{1}{[A]} = 0.033 \times 60 + \frac{1}{[0.0200]_0} \end{equation}
\begin{equation} \frac{1}{[A]} = 1.98 + 50 \end{equation}
\begin{equation} \frac{1}{[A]} = 51.98 \end{equation}
\begin{equation} [A] = \frac{51.98}{1} \end{equation}
[A] = 0.0192

Are the answers I got right? Thanks

[mjc123]

You can't use the same equation for the second part as you did for the first. You have a reaction that is first order in two reagents, i.e.
Rate = k[A][B ]
Now in the first case, the initial concentrations of A and B are the same, and as they react in a 1:1 ratio, [A] = [B ] throughout the reaction. In that case you can write
Rate = k[A]²
and that is where your integrated equation comes from. In the second case the initial concentrations are different, and [A] ≠ [B ] throughout, so you can't use that equation. You have to use the integrated form for the true rate equation above.

[earthnation112]

Really appreciate the reply, do I use the following equation :

\begin{equation}k = \frac{ln(\frac{b}{a}(\frac{a-x} {b -  x}))  }{( a -  b)t } \end{equation}

[mjc123]

If a = [A]₀ and b = [B ]₀; [A] = a-x and [B ] = b-x, then yes.

[earthnation112]

Thank You again for your reply, I want to work out the concentration of sodium ethoxide after 60 mins, I know that:

a = 0.0300
b = 0.0200
t  = 10mins
I want to know the following [A] = a - x
For me to get the answer I desire I require the value "x", how do I obtain it? Do I have to some how rearrange the equation so make x the subject? If so would it be possible to direct me to the right rearranged equation.
Also is the k value the same as the value I obtained for k in the first part of the question?

Thanks again really appreciate the input.

[mjc123]

Yes, it is the same value of k.
Yes, you need to rearrange the equation. You can't do that for yourself?

[earthnation112]

Really appreciate the help, and your right I can rearrange the equation myself. Really appreciate all the help and positive input really means a lot, Thank You