Question

In a first order reaction A⇒2B; the total pressure changes from 100 to 150 mm Hg in 10 minute from the start of the reaction. the rate of reappearance of B is?

Answer 1

Answer:

rate of reappearance of B= 10 mmHg/minute

Explanation:

The rate of disappearance/appearance is the change of product or reactant with respect to time.

The extent of reaction is the change in concentration of  with respect to time divided by it's stoichiometric coefficient.

For a reaction $aA + bB -> cC+ dD$

We define average rate as

Rate= 1/a*(rate of disappearance of A) = 1/b*(rate of disappearance of B)=1/c*(rate of appearance of C)= 1/d *(rate of appearance of D)

since pressure∝ number of moles

we can also take pressure change as rate of reaction

$Rate= -\frac{1}{1}*\frac{change in prseeue of A}{time}=\frac{1}{2}*\frac{change of pressure of B}{time}$

A→ 2B

intial pressure 100 mmHg

final pressure 150 mmHg

initial pressure was due to A only

finally pressure is due to A and B

let's say P pressure got reduced from A then 2P pressure is formed due to B.

We have final total pressure is 150 by the argument

100-P +2P=150

P=50 mm Hg

the pressure due to B will be 100 mm Hg

this pressure formed in 10 mins so

100 mm Hg in 10 mins so 10 mmHg/min

If we put this rate of appearance of B in  $Rate= -\frac{1}{1}*\frac{change in prseeue of A}{time}=\frac{1}{2}*\frac{change of pressure of B}{time}$

we get average rate as 5 mm Hg/min