The value of K_eq for the equilibrium H₂ (g) + I₂ (g) 2HI (g)
Is 794 at 25 °C.At this temperature,what is the value of K_eq for the equilibrium below?
HI (g) 1/2 H₂ (g) + 1/2 I₂ (g)

Question

The value of K_eq for the equilibrium H₂ (g)+ I₂ (g) 2HI (g) Is 794 at 25 °C.At this temperature,what is the value of K_eq for the equilibrium below? HI (g) 1/2 H₂ (g)+ 1/2 I₂ (g)

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: K_eq = ([H₂]^[I₂])/[HI] From the given equilibrium constant value, we can write: 794 = ([H₂]^[I₂])/[HI] For the equilibrium HI (g) 11ea7e7c_6153_ae6a_9a0a_f93371f35b83_TB1194_11 1/2 H₂ (g)+ 1/2 I₂ (g), we can write the equilibrium constant expression as: K_eq = ([H₂]^0.5*[I₂]^0.5)/[HI] In the given equation, the stoichiometric coefficient of HI is 1, and the stoichiometric coefficients of H₂ and I₂ are both 1/2. Therefore, we need to take the square root of the given equilibrium constant value to find the equilibrium constant value for the new equation. K_eq = √794 = 28 Therefore, the answer is D) 0.035.

The Value of Keq for the Equilibrium H2 (G)+ I2

The Value of K_eq for the Equilibrium H₂ (G)+ I₂