A sample of HI(g) is placed in a 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) ⇌ H2(g)+I2(g)

Asked by Pragya Singh | 1 year ago |  77

##### Solution :-

The initial concentration 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)  ⇌   H2(g) +  I2(g)

Initial conc.      0.2 atm       0         0

At equilibrium  0.4 atm,

$$\dfrac{0.16}{2}$$,  $$\dfrac{0.16}{2}$$   =0.08atm

Therefore,

$$K_{p=}\dfrac{p_{H_{2}}\times p_{I_{2}}}{p^{2}_{HI}}\\ \\$$

$$=\dfrac{0.08\times 0.08}{(0.04)^{2}}\\ \\$$

$$=\dfrac{0.0064}{0.0016}\\ \\$$

$$=4.0$$

Hence, the value of Kp for the given equilibrium is 4.0.

Answered by Abhisek | 1 year ago

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