This is just a general question. The integral of the second order reaction is 1/

Question

This is just a general question. The integral of the second order reaction is 1/C-1/C0=2kt on my textbook. However, if I integrate -d[A]/dt = k[A]2 I get 1/C-1/C0=kt instead. Is it Okay? And when I solve problems, do I need tointegrate each time and get what eact equation I need to use? orCan I just use 1/C-1/C0=2kt? Thanks. This is just a general question. The integral of the second order reaction is 1/C-1/C0=2kt on my textbook. However, if I integrate -d[A]/dt = k[A]2 I get 1/C-1/C0=kt instead. Is it Okay? And when I solve problems, do I need tointegrate each time and get what eact equation I need to use? orCan I just use 1/C-1/C0=2kt? Thanks. I get 1/C-1/C0=kt instead. Is it Okay? And when I solve problems, do I need tointegrate each time and get what eact equation I need to use? orCan I just use 1/C-1/C0=2kt? Thanks.

Explanation / Answer

The second order Rate law equation we can writeas:      d[A] / dt = K [A]2         separating thevaribales and integrating the equation with the variables of C0 to C from time t = 0 to t       d [ A ] / [ A ]2 = K dt          = - 1/ A ] c0c   = K t           1 / c -1 /c0 = kt               separating thevaribales and integrating the equation with the variables of C0 to C from time t = 0 to t       d [ A ] / [ A ]2 = K dt          = - 1/ A ] c0c   = K t           1 / c -1 /c0 = kt      

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