A community for students.
Here's the question you clicked on:
 0 viewing
Lena772
 one year ago
If the rate constant of a reaction is 0.33 min1 and the initial reactant concentration is 0.088 M, how many minutes will it take for the reactant concentration to equal 0.013 M?
Lena772
 one year ago
If the rate constant of a reaction is 0.33 min1 and the initial reactant concentration is 0.088 M, how many minutes will it take for the reactant concentration to equal 0.013 M?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0R U sure the units of your rate constant is accurate. It usually includes M

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0That's what the question says :/

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0Can you help me with this one? @timo86m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Need more info then this will do (reactant side) > (products side) In that case dont worry about how many reactants. Just know they produce 2 products CS2 and Cl2. Those reactants produce 3 times as much Cl2 per given time So 3/1

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0May it it be 3/1 ? Cause 3/1 is incorrect. @timo86m

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0It's not D nor C. I only have one more try so I'm not sure whether to chose A or B.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wouldn't chose negatives. The reason could be the equation is not balanced and you have to balance it first :(

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0@Zale101 can you please help me with the original question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0assuming rate = k * [A] .33 min^1 * .088 M = 0.02904 M/min Now u can use the rate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use a simple y = m x + b .013=.02904 x + .088 to get 2.58 y is your wanted M m is the rate x is the minutes and b is original http://www.wolframalpha.com/input/?i=.013%3D.02904+x+%2B+.088 graph

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Hi @lena772 ! Your given rate constant is 0.33 \(\sf min^{1}\), check the unit, it indicates that the reaction is of 1st order!! Consider the reaction to be R > P We can write that Rate = k[R} or \(\sf \Large \frac{d[R]}{dt}\)\(\sf =k[r]^1\\\Large \frac{d[R]}{[R]}=kdt\) Integrating both sides, \(\sf \large\int_{R_0}^{R}\frac{1}{[R]}.d[R]=\large\int_{0}^{t}k.dt\) That gives us, \(\sf ln[R]ln[R_0]=kt\), where [R] is final concentration and \(\sf [R_o]\) is initial concentration of the reactants and t is your time and k is rate constant. Now substituting the values you have, ln[0.013]ln[0.088]= 0.33 \(\times\) t => \(\sf ln\frac{0.013}{0.088}=0.33t\) Solve the equation and you can easily find the value of time. Remember you don't need to do all that calculus stuff each time. You just need to identify the order of reaction by seeing the unit of rate constant and use the integral rate formulas accordingly.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0This is a hint for your second question. Suppose there is a reaction 2A + 1/3B > 6C + 3D Then we can write, Rate = \(\sf \frac{1}{2}\frac{d[A]}{dt}=3\frac{d[B]}{dt}=\frac{1}{6}\frac{d[C]}{dt}=\frac{1}{3}\frac{d[D]}{dt}\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.