Question

Integrated Rate Laws: Working with First Order Reactions When formic acid is heated, it decomposes to give hydrogen and carbon dioxide. This first order decay is shown below: HCOOH (g) → CO2 (g) + H2 (g) At 535.0° C, the half life for this reaction is 23.5 min. How many seconds are needed for the initial concentration of HCOOH to decrease by a factor of 10? seconds

Answer 1

Given reaction is

HCOOH(g) ------> CO2(g) + H2(g) is first order.

Therefore half life=0.693/k

Given half life=23.5 min=23.5 x 60 s=1410 s.

Therefore rate constant, k=0.693/half life=0.693/1410 s=4.915x10^-4 s.

Now the concentration of HCOOH decreacsed by a factor of 10.

The first order rate law is

ln[HCOOH]/[HCOOH]0=- kt

GIVEN [HCOOH]/[HCOOH]0=1/10

ln(1/10)=- (4.915x10^-4 s^-1)t

t=(-2.303)/(-4.915x10^-4 s^-1)

Time=4684.8 s ~ 4685 s.

Please let me know if you have any doubt. Thanks