Question

A gaseous substance dissociates in a rigid vessel at a constant temperature at zone p gives q + r the pressure at different instant is observed as shown calculate the rate constant of dissociation

Answer 1

Question:- A gaseous substance dissociates in a rigid vessel at a constant temperature following first order kinetics, p gives q + r the pressure at different instant is observed as shown:-

Initial Pressure = 1.5atm at time = 0 and Total Pressure = 2atm at time = 20mins.

Calculate the rate constant of dissociation.

Rate constant of dissociation for the given Reaction is 1/20 ln(3/2).

Given:-

p ------> q + r

Order of reaction = 1

Initial Pressure (p0) = 1.5atm

Total Pressure (p) = 2atm

Initial time (t0) = 0

Final time (t) = 20 mins

To Find:-

Rate constant of dissociation for the given Reaction.

Solution:-

To find out the Rate constant of dissociation for the given Reaction, we should follow these simple steps which includes:-

Reaction = p -----> q + r

at t = 0 p0 0 0

at t = 20 p0-x x x

Now,

Total Pressure = p0-x + x + x

Total Pressure = p = 2

$p = p0 - x + x + x$

$p = p0 + x$

$2 = 1.5 + x$

$x = 2 - 1.5$

$x = 0.5$

According to the formula of Rate dissociation of First Order Reaction,

$k = \frac{2.303}{t} log( \frac{po}{po - x} )$

$k = \frac{2.303}{20} log( \frac{1.5}{1.0} )$

$k = \frac{2.303}{20} log( \frac{3}{2} )$

$k = \frac{1}{20} ln( \frac{3}{2} )$

Hence, Rate constant of dissociation for the given Reaction is 1/20 ln(3/2).

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