## Lena772 one year ago If the rate constant of a reaction is 0.33 min-1 and the initial reactant concentration is 0.088 M, how many minutes will it take for the reactant concentration to equal 0.013 M?

1. Lena772

@timo86m

2. anonymous

R U sure the units of your rate constant is accurate. It usually includes M

3. anonymous

or M-1 i mean

4. Lena772

That's what the question says :/

5. Lena772

Can you help me with this one? @timo86m

6. anonymous

Need 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

7. Lena772

May it it be -3/1 ? Cause 3/1 is incorrect. @timo86m

8. anonymous

its not d?

9. Lena772

It's not D nor C. I only have one more try so I'm not sure whether to chose A or B.

10. anonymous

i wouldn't chose negatives. The reason could be the equation is not balanced and you have to balance it first :(

11. Lena772

@Zale101 can you please help me with the original question

12. anonymous

assuming rate = k * [A] .33 min^-1 * .088 M = 0.02904 M/min Now u can use the rate

13. anonymous

You 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

14. Abhisar

Hi @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.

15. Abhisar

This is a hint for your second question. Suppose there is a reaction 2A + 1/3B ----> 6C + 3D Then we can write, Rate = $$\sf \frac{1}{2}\frac{-d[A]}{dt}=3\frac{-d[B]}{dt}=\frac{1}{6}\frac{d[C]}{dt}=\frac{1}{3}\frac{d[D]}{dt}$$