Ch.13 - Chemical KineticsWorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
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Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Phosphine, PH3(g), decomposes according to the equation 4PH3(g) → P4(g) + 6H2 (g) The kinetics of the decomposition of phosphine at 950 K was followed by measuring the total pressure in the system as


Phosphine, PH3(g), decomposes according to the equation 

4PH3(g) → P4(g) + 6H(g) 

The kinetics of the decomposition of phosphine at 950 K was followed by measuring the total pressure in the system as a function of time. The data to the right were obtained in a run where the reaction chamber contained only pure phosphine at the start of the reaction. Choose the rate law that describes this reaction. 

Calculate the value of the rate constant and choose the correct units.

k = 


Part A.

We are asked to determine the Rate Law that describes the decomposition of Phosphine, given the temperature and the total system pressure as a function of time.

The chemical equation is:

4PH3(g)  →   P4(g) + 6H2(g)

Recall that the rate of reaction is the change in pressure (in this case) of a reactant for every change in timeSince we are given with the Total pressure of the system and we are only concerned with the reactant, we need to first identify the Pressure of PHfor every given time.

At t=0, there was only PH3, therefore:

PTotal=PPH3=100 torr 

At t=40 min, we need to use the ICF chart to  determine the pressure of PH3.

             4PH3(g)   →       P4(g)         +       6H2(g)

At t=40 min, total system pressure is 151 torr, therefore, 


   x = 17

Therefore, pressure of PHat t=40 is:

PPH3=100 torr-4xtorrPPH3=100 torr-4(17)torrPPH3=32 torr

At t=40 min, pressure of PH3 is 32 torr.

At t=80 min, total system pressure is 168 torr, therefore, 


   x = 22.7

Therefore, pressure of PHat t=80 is:

PPH3=100 torr-4xtorrPPH3=100 torr-4(22.7)torrPPH3=9.2 torr

At t=80 min, pressure of PH3 is 9.2 torr.

At t=100 min, total system pressure is 171 torr, therefore, 


  x = 23.7

Therefore, pressure of PHat t=100 is:

PPH3=100 torr-4xtorrPPH3=100 torr-4(23.7)torrPPH3=5.2 torr

At t=100 min, pressure of PH3 is 5.2 torr.

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