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  • A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium ? 2HI (g) H_2 (g) + I_2 (g)

7.11     A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

                2HI_{(g)}\rightleftharpoons H_{2}_{(g)}+I_{2}_{(g)}

Answers (1)

best_answer

The initial pressure of HI is 0.2 atm . At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI  is 0.2 - 0.04 = 0.16.

The given reaction is:

2HI_{(g)}\rightleftharpoons H_{2}_{(g)}+I_{2}_{(g)}

HI_{(g)}\rightleftharpoons 1/2H_{2}_{(g)}+1/2I_{2}_{(g)}

At equilibrium,
\\pHI = 0.04\\ pH_2 = \frac{0.16}{2}=0.08\\ pI_2 = \frac{0.16}{2}=0.08

Therefore, 
K_p = \frac{pH_2\times pI_2}{p^2HI}
         \\=\frac{0.08 \times 0.08}{(0.04)^2}\\ =4

Posted by

manish

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