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Question
At 500 K, equilibrium constant, \[\ce{K_c}\], for the following reaction is 5.
\[\ce{1/2 H2 (g) + 1/2 I2 (g) ⇌ HI (g)}\]
What would be the equilibrium constant \[\ce{K_c}\] for the reaction
\[\ce{2HI (g) ⇌ H2 (g) + I2 (g)}\]
Options
0.04
0.4
25
2.5
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Solution
0.04
Explanation:
If the equation is multiplied by 2, the equilibrium constant for the new equation is the square of \[\ce{K}\] it means we must do the square of 5 that is 25. Then and on reversing the reaction the value of the equilibrium constant is inversed means the value of K comes out to be 1/25 = 0.04.
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