
At 800°C,
Calculate K at 8000C for
(a) the synthesis of one mole of H2S from H2 and S2 gases.
(b) the decomposition of one mole of H2S gas.

(a)
Interpretation:
The equilibrium constant for the formation of 1 mol of
Concept introduction:
The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.
For a general reaction as follows:
The expression for the equilibrium constant is represented as follows:
Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.
Answer to Problem 16QAP
The equilibrium constant for the synthesis of 1 mol of
Explanation of Solution
The given equilibrium reaction is as follows:
The value of equilibrium constant at
The expression for the equilibrium constant will be:
Thus,
Now, the reaction for the synthesis of 1 mol of
The expression for the equilibrium constant of the above reaction will be:
Since,
Taking reciprocal of the above equation:
Now, taking square root:
Therefore, equilibrium constant for the synthesis of 1 mol of

(b)
Interpretation:
The equilibrium constant for the decomposition of 1 mol of
Concept introduction:
The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.
For a general reaction as follows:
The expression for the equilibrium constant is represented as follows:
Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.
Answer to Problem 16QAP
The equilibrium constant for the decomposition of 1 mol of
Explanation of Solution
The given equilibrium reaction is as follows:
The value of equilibrium constant at
The expression for the equilibrium constant will be:
Thus,
The reaction for the decomposition of 1 mol of
The expression for the equilibrium constant will be:
Since,
Taking square root:
Therefore, the equilibrium constant for the decomposition of 1 mol of
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Chapter 12 Solutions
Chemistry: Principles and Reactions
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