Question

A reaction that is first order with respect to reactant A has a rate constant 6 min^-1. If we start with [A]=50mol/L, when would [A] reach the value 0.05mol/L.

Answer 1

Just insert into the equation. You have all the information:
\[[A]=[A _{0}]e ^{-kt}\]
Isolate t:
\[t=\frac{ -\ln(\frac{ [A] }{ [A _{0}] }) }{ k }\]

Answer 2

Do a check for the units your self ;)

Answer 3

plz solve using \[t=\frac{ 2.303 }{ k }\log_{10}\frac{ A_{0} }{ A } \]

Answer 4

Also a option... just insert into the equation.

Answer 5

ya so what is A and A0 here.......i am not getting the answer i request you to solve the question step by step and show the solution. I have tried doing the sum 5 times can u plz help me.......?

Answer 6

Okay. I help you the 0 in A0 is a reference to the following A at t=0... so A0 = 50mol/L.... A=0.05mol/L. k is rate constant and so on... in other words:
\[t=\frac{ -\ln(\frac{ [A] }{ [A _{0}] }) }{ k }=\frac{ -\ln(\frac{ [0.05] }{ 50 }) }{ 6 }\
I've been lazy and not put the units on the calculations... you should do so.

Answer 7

\[t=\frac{ -\ln(\frac{ [A] }{ [A _{0}] }) }{ k }=\frac{ -\ln(\frac{ [0.05] }{ 50 }) }{ 6 }\]

Answer 8

I am Sorry but after substituting the values we get 0.5min as the answer but the solution says it is 0.77min. Please check......

Answer 9

I don't see how you get 0.5... nor 0.77 I want it to be 1 min and 15 sec.
\[t=\frac{ -\ln(\frac{ [0.05] }{ 50 }) }{ 6 }=\frac{ -\ln (0.001) }{ 6 } \approx \frac{ -(-6.90776) }{ 6 } \approx 1.15129\]

Answer 10

sorry not 1 min and 15 sec but 1.15 min.

Answer 11

yaaa its 1.15 :) i too checked thrice

Answer 12

hey please check the question is it 0.05 or 0.5 mol/L
if we take 0.5 ur desired 0.77 min answer we get:)

Answer 13

@leena1996

Answer 14

Alright...thanks..........@chmvijay @Frostbite

Answer 15

Nice noticed chmvijay.

Answer 16

humnn thank you @Frostbite and @leena1996 :)

Answer 17

Well back to electrochemistry for me... I get too easy distracted.

Answer 18

:)

Answer 19

@Frostbite Oops sorry.....get back to Work....!

Answer 20

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