Suppose that Trial I in this lab was run using 0.10 M NaOH, producing a linear kinetics graph with best-fit equation y = -0.00275 x + 0.222. If Trial Il was run using 0.20 M NaOH, what would the slope of the same graph be if the reaction were (a) zero order, (b) first order, or (c) second order?

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rafikiedu08 rafikiedu08 · Answer 1 · 2022-09-16T07:45:39+02:00

For different order Kinetics the lineer graph is different plots.

For Zero order reaction, the integrated rabe law is

[A] = [A]₀-Kt K =rate constant

[A] vs t graph is straight line with slope = -k

For first order reaction, the integrated rate law is

In [A] = In [A]o - Kt

In [A] vs t graph is straight line with slope = -K

For second order reaction, the integrated rate law is

1/[A]= 1/ [A]o + Kt

1/[A] vs t graph is straight line with slope = +k

So, for all cases the slope of straight line is either +k or -K.

k'is rate constant only depends on temperature not on concentration of reactant.

So the slope will not change for Trial II with 0.2m Naoh.

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