From graphs A): determine the reaction order for p (respect to iodide )and q(res

Question

From graphs

A): determine the reaction order for p (respect to iodide )and q(respect to hydrogen peroxide) Obtain the equation for the best straight line. (The slope equals the order of reaction)

B) Determine rate constant for each trial by plugging in the values for p and q into the rate equation for each kinetic trial

log(H20210 log Amol At Trial 7 -1.70E+00 -7.36E+00 Trial 1 -1.52E+00 -7.00E-00 Trial S -6.93E+00 Trial 6 -1.15E-00 -7.17E-00 slope 362617816 Intercept 6 SS98028ss -6.40E-00 5808.00 .700E-00 7.20E-00 determination of q (order with respect to H202) o 3626 S99s R? 0.2029 Series

Explanation / Answer

A) This part is very simple. The order of the reaction is nothing but the slope of the line. Hence reaction order for p (respect to iodide ) is 0.17 and q(respect to hydrogen peroxide) is 0.36.

B) IN econd part in order to find the individual values of rate constant we have to first write the rate law expression.

for p rate law will be d[I]/dt = k[I]0.17

now in this expression we need d[I]/dt and [I] to get k. We are given log(d[I]/dt) value. So taking anti log we get the required value. Similarly for [I] also. Then keep on putting these values in the above rate law expression and you will different k.

Repeat the same procedure for q also.

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