   # The rate law for the decomposition of phosphine (PH 3 ) is Rate = − Δ [ PH 3 ] Δ t = k [ PH 3 ] It takes 120. s for 1.00 M PH 3 to decrease to 0.250 M . How much time is required for 2.00 M PH 3 to decrease to a concentration of 0.350 M ? ### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
Publisher: Cengage Learning
ISBN: 9781305957404

#### Solutions

Chapter
Section ### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
Publisher: Cengage Learning
ISBN: 9781305957404
Chapter 12, Problem 53E
Textbook Problem
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## The rate law for the decomposition of phosphine (PH3) is Rate   =   − Δ [ PH 3 ] Δ t   =   k [ PH 3 ] It takes 120. s for 1.00 M PH3 to decrease to 0.250 M. How much time is required for 2.00 M PH3 to decrease to a concentration of 0.350 M?

Interpretation Introduction

Interpretation: The rate law equation and the time for the change in concentration for the decomposition of phosphine (PH3) is given. By using these values, the time required for the given change in concentration of PH3 is to be calculated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

To determine: The time taken by 2.00M (PH3) to decrease to a concentration to 0.350M .

### Explanation of Solution

Given

The time taken for 1.00M (PH3) to decrease to a concentration of 0.250M is 120s .

The formula of rate of reaction is,

Rate=Δ[PH3]Δt=k[PH3]

Where,

• k is rate constant.
• Δ[PH3] ischange in concentration of [PH3] .
• Δt ischange in time.

The given equation belongs to first order reaction. Hence,

[PH3]0=1.00M[PH3]=0.250M

Where,

• [PH3]0 is initial concentration.
• [PH3] is concentration at time 120s .

Formula

The integral rate law equation of first order reaction is,

kt=ln([A]0[A]) (1)

Where,

• k is rate constant.
• t is time.

Substitute the values of [PH3]0,[PH3] and t in the above equation.

kt=ln([PH3]0[PH3])k×120s=ln(1

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