At 500 K in the presence of a copper surface , ethanoldecomposes according to the equation
C2H5OH(g)-------------CH3CHO(g) + H2 (g)
the pressure of C2H5OH was measured as a function of time andthe following data were obtained:
Time(s) P (C2H5OH) (torr)
0 250.
100. 237
200. 224
300. 211
400. 198
500. 185
since the pressure of a gas is directly proportional to theconcentration of gas, we can express the rate law for a gaseousreactionin terms of partial pressures. using the above data, deducethe rate law, the integrated rate law, and the value ofrate constant, all in terms of pressure units in atm and timein seconds.
predict the pressure of C2H5OH after 900 s from the start ofthe reaction.
(hint: to determine the order of the reaction with respect toC2H5OH, compare how the pressure of C2H5OH decreases with each timelisting.)

Respuesta :

Answer:

Explanation:

rate of reaction

= -ve change in pressure of ethanol / time

= - (250 -237 )/100 = - 13 / 100 torr/s

= - 0.13 torr/s

next

- (237 - 224 )/100 = - 13 / 100 torr/s

= - .13 torr/s

next

- (224 - 211 )/100 = - 13 / 100 torr/s

= - .13 torr/s

so on

So rate of reaction is constant and it does not depend upon concentration or pressure of reactant .

So order of reaction is zero.

rate of reaction =K  [C₂H₅OH]⁰

K is rate constant

K = .13 torr/s

In 900 s decrease in pressure

= 900 x .13 = 117

So after 900s , pressure of ethanol will be

250 - 117 = 133 torr