Question

The rate law for the reaction below is given by Rate = k(P ). At 680°C, the half-life is 35.0 s. Given initial pressures of P = 0.67 atm, P = 0.75 atm, and P = 0.25 atm, how many seconds will it take for the pressure of PH to drop to 0.50 atm

4 PH (g) → P (g) + 6 H (g)

Answer 1

Solution :-

Reaction is first order with P

So the

Rate constant K = 0.693 / t1/2

                             = 0.693 / 35.0 s

                           = 0.0198 s-1

Now lets calculate the time needed to decrease the pressure from 0.67 to 0.50 atm

Ln[P2]/[P1] = - k*t

Ln [0.50 ]/[0.67] = -0.0198 s-1 *t

-0.29267 = -0.0198 s-1 * t

-0.29267 / -0.0198 s-1 = t

14.8 s = t

So the time needed is 14.8 sec

Now lets calculate the time needed to decrease the pressure from 0.75 to 0.50 atm

Ln[P2]/[P1] = - k*t

Ln [0.50 ]/[0.75] = -0.0198 s-1 *t

-0.40547 = -0.0198 s-1 * t

-0.40547 / -0.0198 s-1 = t

20.5 s = t

So the time needed is 20.5 sec