Question

In an autocatalytic chemical reaction, the product formed is a catalyst for the reaction. If Qintial is the amount of the original substance and x is the amount of catalyst formed, the rate of chemical reaction is dQ/dx=kx(Qinitial-x).
For what value of x will the rate of chemical reaction be greatest?

Answer 1

Qinitial is \[Q _{0}\]

Answer 2

k is a constant?

Answer 3

well here what my teacher gave me so far:
\[\frac{ dQ }{ dx }= K _{x} (Q _{0} - x)\]
Find max \[\frac{ dQ }{ dx }\]
\[\frac{ dQ }{ dx } = kQ _{0}x-kx^2\]
\[\frac{ d^2Q }{ dx^2 } = \]

Answer 4

So we want to maximize that rate of change, so we need to differentiate dQ/dx and set it to zero.

Answer 5

yepp

Answer 6

So when you differentiate dQ/dx, and set it to zero, you'll get

\[0 = kQ _{0} - 2kx\]

since k and Q are just constants.

Answer 7

no, make \[\frac{ d^2Q }{ dx^2 } = 0\]

Answer 8

Yes, but i've already differentiated dQ/dx which gives \[\frac{ d ^{2}Q }{ dx ^{2} }\]

I just didn't write in that part, easier to just set it to zero, because that's we'll do anyway (it''s just tedious to write in with the equation editor)

Divide both sides by k, so
\[0 = Q _{0} - 2x\]
add 2x to both sides:
\[2x = Q _{0}\]
so
\[x = \frac{ Q _{0} }{ 2 }\]

Answer 9

and thats my answer?

Answer 10

yes no?

Answer 11

Yep, that's where the rate of change will be greatest, which seems to make sense chemically because, at that point, there's an equal amount of catalyst and remaining substance (since if half the initial amount is now catalyst, then the other half of the initial amount remains).

Beyond that point, you have excess catalyst and not enough substance to make use of it, and before that point, you have excess substance and not enough catalyst.

Answer 12

ok ty :P can you help me with another problem plsss?? <333

Answer 13

sure