A gaseous reaction 2X(g) → Y(g) is observed to be second order with respect to reactant X. If the half-life of this reaction is 20 minutes when the initial partial pressure of X is 0.4 atm, what will be the half-life when the initial partial pressure of X is increased to 0.8 atm, assuming the temperature remains constant?
(A) 5 minutes
(B) 10 minutes
(C) 20 minutes
(D) 40 minutes
Answer (2)
The half-life of the reaction when the initial partial pressure of X is increased to 0.8 atm is 40 minutes. This is determined by knowing that for a second-order reaction, the half-life is inversely related to the initial concentration. As the pressure doubles, the half-life also doubles, resulting in a new half-life of 40 minutes.
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In this problem, we are dealing with a chemical reaction where 2 moles of X(g) react to form 1 mole of Y(g), and it is specified that the reaction is second order with respect to the reactant X.
For a reaction that is second order with respect to a reactant, the half-life t 1/2 is given by the formula:
t 1/2 = k [ A ] 0 1
Where:
t 1/2 is the half-life
k is the rate constant of the reaction
[ A ] 0 is the initial concentration or partial pressure of the reactant
From the problem, the half-life ( t 1/2 ) is 20 minutes when the initial partial pressure [ X ] 0 is 0.4 atm. Therefore, we can use this information to solve for the rate constant k :
20 = k ( 0.4 ) 1
Solving for k :
k = 20 × 0.4 1 = 8 1 = 0.125 atm − 1 min − 1
Next, we need to determine the new half-life t 1/2 when the initial pressure of X is increased to 0.8 atm.
Using the same half-life formula:
t 1/2 = k [ X ] 0 1
Plug in the known values:
t 1/2 = 0.125 ( 0.8 ) 1 = 0.1 1 = 10 minutes
Therefore, the new half-life when the initial partial pressure of X is 0.8 atm is 10 minutes.
The correct multiple-choice option is (B) 10 minutes .
