For a reaction - H - 2 - - - I - 2 - rightleftharpoons 2HI 721 K- the value of equilibrium constant is 50- If 0-5 moles each of - H - 2 - and - I - 2 - is added to the system the value of equilibrium constant will be40305060

Question

For a reaction  - H - 2 -  -  - I - 2 -   rightleftharpoons  2HI 721 K- the value of equilibrium constant is 50- If 0-5 moles each of  - H - 2 -  and  - I - 2 -  is added to the system the value of equilibrium constant will be40305060

Toppr Admin · Accepted Answer

The equilibrium reaction is H2-g-I2-g-x2192-2HI-g-xA0-The relationship between Kp-xA0- and Kc-xA0- -xA0-is Kp-Kc-RT-x394-n -xA0-For the equilibrium reaction- -x394-n-2-x2212-1-1-0-xA0-Hence- Kp-xA0-Kc -x200B-xA0- -50-

For a reaction $H_{2}$ + $I_{2}$ $⇌$ 2HI at 721 K, the value of equilibrium constant is 50. If 0.5 moles each of $H_{2}$ and $I_{2}$ is added to the system the value of equilibrium constant will be
40
30
50
60

The equilibrium reaction is $H_{2} (g) + I_{2} (g) \to 2 H I (g)$ .
The relationship between Kp and Kc is $K p = K c (R T)^{Δ n}$ .
For the equilibrium reaction, Δn=2−(1+1)=0.
Hence, Kp =Kc =50.

For a reaction H2 + I2 ⇌ 2HI at 721 K, the value of equilibrium constant is 50. If 0.5 moles each of H2 and I2 is added to the system the value of equilibrium constant will be40305060

The equilibrium reaction is H2(g)+I2(g)→2HI(g). The relationship between Kp and Kc is Kp=Kc(RT)Δn . For the equilibrium reaction, Δn=2−(1+1)=0. Hence, Kp =Kc ​ =50.

For a reaction $H_{2}$ + $I_{2}$ $⇌$ 2HI at 721 K, the value of equilibrium constant is 50. If 0.5 moles each of $H_{2}$ and $I_{2}$ is added to the system the value of equilibrium constant will be
40
30
50
60

The equilibrium reaction is $H_{2} (g) + I_{2} (g) \to 2 H I (g)$ .
The relationship between Kp and Kc is $K p = K c (R T)^{Δ n}$ .
For the equilibrium reaction, Δn=2−(1+1)=0.
Hence, Kp =Kc =50.