Question

Luigi
1 mol each of H2 and 12 are heated in a closed vessel till equilibrium is obtained. If 40% of 12 is converted to HI, the value of Kc at the
same temperature will be
H2(g) +1,(9)=2HI(g)​

Answer 1

Answer:

Mark me as brainlist answer

Answer 2

Given info : 1 mol of each of H₂ and I₂ are heated in a closed vessel till equilibrium is obtained. if 40% of I₂ is converted to HI.

To find : the value of Kc at the same temperature is ..

solution : association reaction of H₂ and I₂ is..

H₂ (g) + I₂ (g) ⇒2HI (g)

at t = 0 1 1 0

at t = t₀. 1 - α 1 - α 2α

here α = 40% = 0.4

so, [H₂] = 1 - 0.4 = 0.6

[I₂] = 1 - 0.4 = 0.6

[2HI] = 2 × 0.4 = 0.8

now equilibrium constant, Kc = [2HI]²/[H₂][I₂]

= (0.8)²/(0.6)(0.6)

= 0.64/0.36

= 1.7

Therefore the equilibrium constant at the same temperature is 1.7

also read similar questions : Calculate the enthalpy change for the reaction

H2(g)+ Br2(g)2HBr(g)Given :Bond enthalpy of H—H = 435 kJ mol-1Bond enthal...

https://brainly.in/question/2928687

the value of the equilibrium constant for the reaction H2(g)+I2(g)2HI(g) at 720K is 48. What is the value of the equilib...

https://brainly.in/question/1000595