Question

Kp for the reaction H2 + I2 = 2HI at 460° C is 49. if the partial pressures of H and I is 0.5 atm respectively,determine the partial pressure of each gas in the equilibrium pressure.

Answer 1

Answer: The partial pressure for H2 is 0.1112 atm and HI is 0.7776 atm.

Explanation:

                   $H_{2}+ I_{2} \rightleftharpoons 2HI$

Initial:       0.5      0.5         0

Final:   (0.5 - x)  (0.5 - x)    2x

The formula for $K_{p}$ for the reaction  $H_{2} + I_{2}\rightleftharpoons 2HI$ is as follows.

                    $K_{p} = \frac{P_{HI}}{P_{H}_{2}P_{I}_{2}}$

                    $= \frac{2x^{2}}{(0.5-x)(0.5-x)}$

                   $49=\frac{(2x)^{2}}{(0.5-x)^2}$

                   49 x^2 - 49x +12.25 = 4x^2

                    45 x^2 - 49x +12.25 = 0

Now, using the quadratic formula solve for x. We get x = 0.7 or x = 0.3888.

The value x = 0.3888 is valid as the partial pressure of $H_{2}$ and $I_{2}$ is 0.5 atm.

Therefore, $P_{H_2}$ at equilibrium = $P_{I_2}$ = 0.5-x

                       = 0.5 - 0.3888

                       = 0.1112 atm

$P_{HI}$ at equilibrium = 2x = $2 \times 0.3888$

                      = 0.7776 atm

Thus, the partial pressure for H2 is 0.1112 atm and HI is 0.7776 atm.

                 

             

Answer 2

Answer:

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