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Sagot :
Answer: The value of [tex]K_{c}[/tex] for this reaction is 250000.
Explanation:
The given equation is as follows.
[tex]2NH_{3}(g) + 3I_{2}(g) \rightleftharpoons N_{2}(g) + 6HI(g)[/tex]
[tex]N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g); K_{c_{1}} = 0.50[/tex] ... (1)
[tex]H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g); K_{c_{2}} = 50[/tex] ... (2)
To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.
[tex]3H_{2}(g) + 3I_{2}(g) \rightleftharpoons 6HI(g); K_{c_{2}} = (50)^{3}[/tex] ... (3)
Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.
[tex]3I_{2} - N_{2} \rightleftharpoons 6HI - 2NH_{3}[/tex]
This equation can also be re-written as follows.
[tex]3I_{2} + 2NH_{3} \rightleftharpoons N_{2} + 6HI[/tex]
This equation is similar to the equilibrium equation given to us.
Therefore, during this subtraction the equation constants get divided as follows.
[tex]K^{'}_{c} = \frac{K_{c_{2}}}{K_{c_{1}}}\\= \frac{(50)^{3}}{0.50}\\= 250000[/tex]
Thus, we can conclude that the value of [tex]K_{c}[/tex] for this reaction is 250000.
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