CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The Kp value for the reaction $$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$$ at $$460^oC$$ is 49. If the initial pressure of $$H_2$$ and $$I_2$$ are 0.5 atm respectively, determine the partial pressure of $$H_2, I_2$$ and $$HI$$ gases at equilibrium.


A
0.11 atm, 0.11 atm, 0.77 atm
loader
B
0.11 atm, 0.77 atm, 0.11 atm
loader
C
0.77 atm, 0.11 atm, 0.11 atm
loader
D
None of these
loader

Solution

The correct option is A 0.11 atm, 0.11 atm, 0.77 atm
                     $$H_2 \ \ \ \ \ \           + \ \ \ \ \  I_2   \  \ \ \  \to \ \ \ \ \ \ 2HI$$

Initial     :          $$ 0.5$$                $$0.5$$               $$ 0$$
At eq. :       $$(0.5−x)$$     $$(0.5−x)$$          $$2 x$$

$$ K_p = \dfrac{(2x)^2}{(0.5−x)^2}   = 49$$

On solving the above equation, $$2x= 3.5−7x$$

$$9x =3.5$$

$$x = 0.389$$

At equillibrium,Partial pressure of hydrogen, $$p'H_2 = 0.5− 0.389 = 0.111\ atm$$

Partial pressure of $$I_2, p'I_2 = 0.5− 0.389 = 0.111\ atm$$

Partial pressure of hydrogen iodide , $$p'HI = 2\times  0.389 =0.77\ atm$$

Hence, option A is correct.

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image