(a)
Interpretation:
The equilibrium partial pressures of
(a)
Explanation of Solution
The given reaction is shown below,
The reaction can be expressed as follow,
The reaction tells that two molecule of phosphine gives two molecule of phosphorus and three molecules of water.
ICE table:
Pressure
Initial concentration | 0 | 0 | |
Change | |||
At equilibrium |
Pressure,
The equilibrium partial pressures of
The equilibrium partial pressures of
(b)
Interpretation:
Once equilibrium is reached, the mass of phosphorus produced has be calculated.
(b)
Explanation of Solution
The equilibrium partial pressures of
Number of moles of phosphorous is calculated as follows,
Mass in grams of phosphorous is calculated as follows,
Mass in grams of phosphorous is
(c)
Interpretation:
Equilibrium constant K has be calculated.
(c)
Explanation of Solution
The given reaction is shown below,
The reaction can be expressed as follow,
The reaction tells that two molecule of phosphine gives two molecule of phosphorus and three molecules of water.
ICE table:
Pressure
Initial concentration | 0 | 0 | |
Change | |||
At equilibrium |
Pressure,
The equilibrium partial pressures of
The equilibrium partial pressures of
Equilibrium constant K is calculated as follows,
Equilibrium constant K is 183.
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Chapter 5 Solutions
Chemical Principles: The Quest for Insight
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