Question

2.00 mol of H2(g) and 1.00 mol of I2(g) are placed in a 1.00 L container, and they react to form HI(g). At equilibrium, it is found that 1.80 moles of HI(g) are present in the container. Calculate K for the reaction: H2(g) I2(g) ⇄ 2 HI(g)

Answer 1

The answer to your question is: k = 1.62

Step-by-step explanation:

Data

H2(g) = 2.00 mol

I2 (g) = 1.00 mol

HI (g) = 1.80 mol

K = ?

Reaction H2(g) + I2(g) ⇄ 2 HI(g)

Formula

`k = ([[HI]^(2) )/([H2][I2])`

Substitution

`k = ([1.8]^(2) )/([2][1])`

k =
`(3.24)/(2)`

k = 1.62