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

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

0

Similar questions
View More

People also searched for
View More