Question

At 1800 K ethane gas decomposes to ethene
and hydrogen. Rate constant for the reaction is
10-3pa 1sec-1. If initial pressure of ethane is
3x10 Pa, how many sec. would it take for the
pressure to reach 5 x 10^5Pa?
​

Answer 1

at 1800K, ethane gas decomposes to ethene and hydrogen. rate constant for the reaction is 10¯³ Pa¯¹sec¯¹. initial pressure of ethane is 3 × 10^5 Pa.

To find : time required to reach the pressure 5 × 10^5 Pa.

solution : ethane gas decomposes to ethene and hydrogen. it is a second order reaction.

i.e., rate, r = k[A]²

-d[A]/dt = k[A]²

⇒∫d[A]/[A]² = -∫dt

⇒-1/[A] + 1/[A]₀ = -kt

⇒1/[A] = 1/[A]₀ + kt ....this is required expression.

as reaction is C2H6(g) => C2H4(g) + H2(g)

at t = 0, 3 × 10^5 Pa 0 0

at t = t, (3 × 10^5 - x ) x X

given, total pressure = 5 × 10^5 Pa

(3 × 10^5 - x) + x + x = 5 × 10^5

⇒x = 2 × 10^5

so, pressure of C2H6 at time t = 3 × 10^5 - 2 × 10^5 = 10^5 Pa

now initial pressure of C2H6 , [A]₀ = 3 × 10^5 Pa

final pressure of C2H6 , [A] = 10^5 Pa

k = 10^-3 Pa^-1 sec^-1

so, 1/(10^5) = 1/(3 × 10^5) + (10^-3) × t

⇒2/3 × 10^-5 = 10^-3 × t

⇒t = 0.67 × 10^-2 sec = 6.7 ms

Therefore the time required to reach the pressure 5 × 10^5 Pa is 6.7 ms

Answer 2

Answer:

6.7 ms

Explanation:

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