Can you guys pls help me with this

Theoretical yield of hydrogen iodide = 552.96 grams

Percentage yield = 83%

Explanation:

Hydrogen ar- 1

Iodine ar- 127

Theoretical yield of hydrogen iodide =

4.32/1 = 4.32 moles of hydrogen

4.32 * 127 = 548.64 grams of iodine

4.32 + 548.64 = 552.96 grams of hydrogen iodide

Percent yield = 460/552.96 = 0.83

0.83 * 100 = 83%

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

answer: i would say b.

explanation: