Question 2 The following reaction: was studied by measuring the [X] as a functio

Question

Question 2 The following reaction: was studied by measuring the [X] as a function of time. Reactant Y was always present in excess and it was assumed that the [Y] remained essentially constant during the reaction Shown below is a table of the [X] as a function of time, in seconds. In the first experiment, [Y] = 2.0 M and in the second experiment, [Y] = 4.0 M. Also shown are the plots of these data which gave linear relationships [Y] = 2.0 M [Y] = 4.0 M t (s) [X] (M) [x] (M) 1.000 0.916 0.837 0.765 0.700 0.837 0.700 0.548 0.488 0.407 2 4 6 0.05 0.00 0.05 0.2 0.3 -0 .20 0.25 0.30 0.8 0.9 0.35 0.40 time (s) time (s) Use these data to determine the form of the rate equation for this reaction The experimental rate equation has the form: rate = k [X] TY] Submit Answer Tries 0/99

Explanation / Answer

since, plot between ln[X] and time is straight line with negative slope
so, rate is first order with respect to X

slope of first plot = (-0.9 + 0.2)/(6-2)
= -0.175
slope of second plot = (-0.35 + 0.00)/(6-2)
= -0.0875

slope become half when we double the [Y]
so, order is 1 with resspect to Y

so,
rate = k*[X]^1*[Y]^1

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