Sapling week 9/10 Number 5

[Phoebe Joseph]

Could someone walk me through their logic on this question? I got the right answer but I just want to make sure I'm not making any wild assumptions to get to my answers... Thank you so much!!!

[Can Yilgor 2D]

According to the rate law, the reaction is first‑order in A and second‑order in B. If the concentration of a first‑order reactant is halved, then the reaction rate changes by a factor of 1/2. If the concentration of a second‑order reactant is tripled, then the reaction rate changes by a factor of (3)^2=9. Therefore the overall reaction rate changes by a factor of 1/2×9.

Similarly, if the concentration of a first‑order reactant is tripled, then the reaction rate changes by a factor of 3. If the concentration of a second‑order reactant is halved, then the reaction rate changes by a factor of (1/2)^2=1/4. Therefore, the overall reaction rate changes by a factor of 3×(1/4).

[Phoebe Joseph]

Thank you so much!

[Rajshree 1F]

depending on the question, you add the integers needed in the concentration brackets and if that concentration is raised to a power, you also raise that integer to that value. after that multiply the initial rate given by the integers u have. if A is halved and B is tripled it would be rate = k[.5A][3B]^2. u would square 3 and therefore get rate = 9/5*k[A][B]^2 and u would substitute k[A][B]^2 with the initial rate and multiply it with 9/2.

[Daniela Santana 2L]

Hi! The way that I thought about solving this problem was by adding the 1/2 and 3 coefficient where the problem said it should go. I then set up the rate equation that was also given with the new coefficients in their respective places. I multiplied the coefficients together then multiplied this value by the initial rate given. I hoped this helped!