Okay, here is another problem that I am pulling my hair outwith. A first-order r

Question

Okay, here is another problem that I am pulling my hair outwith. A first-order reaction has a rate constant of 3.0 x 10e-3/s.The time required for the reaction to be 75% complete is_____? Using the same book I am following example 13.4, (c).
This is what my equation looks like thus far, but the answers I amgetting are not even close that what is offered. Ugh! Equation: ln [At] / [A0] = -kt   this equationcan be arranged to look like this too: ln [A]t = -kt + ln[A]0 ln (.25/1.0) = (3.0x10e-3)t To find the time required, would I not take 2.5 and divide itby the constant of .003? Just a general statment here, but I am 26 and feel like I amgong to stroke out because of this class... The example that is given in the book has this: ln(.26/1.0) = -(6.7x10e-4se-1)t then they have t = 2.0 x 10e3s x (1 min/ 60s) = 33 min How in the world are they getting 2.0 x 10e3s? Someone please help me! I am so confused.... Again. Okay, here is another problem that I am pulling my hair outwith. A first-order reaction has a rate constant of 3.0 x 10e-3/s.The time required for the reaction to be 75% complete is_____? Using the same book I am following example 13.4, (c).
This is what my equation looks like thus far, but the answers I amgetting are not even close that what is offered. Ugh! Equation: ln [At] / [A0] = -kt   this equationcan be arranged to look like this too: ln [A]t = -kt + ln[A]0 ln (.25/1.0) = (3.0x10e-3)t To find the time required, would I not take 2.5 and divide itby the constant of .003? Just a general statment here, but I am 26 and feel like I amgong to stroke out because of this class... The example that is given in the book has this: ln(.26/1.0) = -(6.7x10e-4se-1)t then they have t = 2.0 x 10e3s x (1 min/ 60s) = 33 min How in the world are they getting 2.0 x 10e3s? Someone please help me! I am so confused.... Again.

Explanation / Answer

Okay, here is another problem that I ampulling my hair out with. A first-order reaction has a rate constantof 3.0 x 10e-3/s. The time required for the reaction to be 75%complete is _____? Using the same book I am following example13.4, (c).
This is what my equation looks like thus far, but the answers I amgetting are not even close that what is offered. Ugh! Equation: ln [At] / [A0] =-kt   this equation can be arranged to look like thistoo: ln [A]t = -kt + ln[A]0 ln (.25/1.0) =(3.0x10e-3)t To find the time required, would I not take2.5 and divide it by the constant of .003?

No, you take the natural log of 0.25 and divide by 3e-3.

the natural log is taking the logarithm with base e, where e ~2.71.... (Are you familiar with logarithms of base 10 and 2?)

ln(0.25/1) = -1.386 = -k*t   (you forgot the negativesign in your expression :ln (.25/1.0) = (3.0x10e-3)t

Thus t = -1.386/(-3.0x10^-3) = 462 second = 7.7

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ln(.26/1.0) = -(6.7x10e-4se-1)t then they have t = 2.0 x 10e3s x (1 min/ 60s) = 33min How in the world are they getting 2.0 x10e3s? Someone please help me! I am so confused....Again.
The 2.0 x 10^3 is obtained

ln(0.26/1.0) = -1.347     

http://www.google.com/search?q=ln(0.26%2F1.0)&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

ln(.26/1.0) = -(6.7x10e-4se-1)t then they have t = 2.0 x 10e3s x (1 min/ 60s) = 33min How in the world are they getting 2.0 x10e3s? Someone please help me! I am so confused....Again.

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