Question

. At 400 K, the half-life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 28.9 kPa, the half-life as 178 s. Determine the order of the reaction.

Answer 1

Answer:

The order of this half-life reaction is zero. This is a zero order reaction.

Explanation:

Step 1: Data given

At the pressure P0,1 of 55.5 kPa, the half-life is 340 s

At the pressure P0,2 of 28.9 kPa, the half-life is 178 s

Step 2: Calculate the order

(n-1) = ((log (t1/2P0,1 / t1/2P0,2)) / (log (P0,2 /P0,1)))

⇒ with t1/2P0,1 = the half-life at a pressure of 55.5 kPa = 340 s

⇒ with t1/2P0,2 = the half-life at a pressure of 28.9 kPa = 178 s

⇒ with P0,1 = the pressure of 55.5 kPa

⇒ with P0,2 = the pressure of 28.9 kPa

(n-1) = (log (340/178)) /  log( 28.9/55.5)

(n-1) = log (1.91) / log(0.52)

(n-1) = 0.281 / -0.284

n-1 = -1

n = 0

The order of this half-life reaction is zero. This is a zero order reaction.