How to calculate the concentration of a reactant after a specific time with the rate given for a second order reaction?
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If you are a high school student, you likely do not need to know the derivation below. Typically, this formula is memorized:
1[NOBr]=1[NOBr]0+k⋅t1[NOBr]=1[NOBr]0+k⋅t
For derivation of the formula above, the disappearance of [NOBr][NOBr] is second order, so we write this out as a differential equation:
−d[NOBr]dt=k⋅[NOBr]2−d[NOBr]dt=k⋅[NOBr]2
We rearrange all [NOBr][NOBr] terms on the left, and all time terms on the right:
−1[NOBr]2⋅d[NOBr]=k⋅dt−1[NOBr]2⋅d[NOBr]=k⋅dt
Then, this solved by integration of both sides:
1[NOBr]−1[NOBr]0=k⋅t1[NOBr]−1[NOBr]0=k⋅t
Then, move the initial [NOBr]0[NOBr]0 term over using algebra:
1[NOBr]=1[NOBr]0+k⋅t1[NOBr]=1[NOBr]0+k⋅t
With this form, you enter your information to get the answer simply. The method of solving differential equations can be used for other rate laws.
1[NOBr]=1[NOBr]0+k⋅t1[NOBr]=1[NOBr]0+k⋅t
For derivation of the formula above, the disappearance of [NOBr][NOBr] is second order, so we write this out as a differential equation:
−d[NOBr]dt=k⋅[NOBr]2−d[NOBr]dt=k⋅[NOBr]2
We rearrange all [NOBr][NOBr] terms on the left, and all time terms on the right:
−1[NOBr]2⋅d[NOBr]=k⋅dt−1[NOBr]2⋅d[NOBr]=k⋅dt
Then, this solved by integration of both sides:
1[NOBr]−1[NOBr]0=k⋅t1[NOBr]−1[NOBr]0=k⋅t
Then, move the initial [NOBr]0[NOBr]0 term over using algebra:
1[NOBr]=1[NOBr]0+k⋅t1[NOBr]=1[NOBr]0+k⋅t
With this form, you enter your information to get the answer simply. The method of solving differential equations can be used for other rate laws.
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