The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) is at equilibrium at 543K. Before the reaction, the concentrations of the gases were [SO2] = 0.060 mol/L, [O2] = 0.050 mol/L, and [SO3] = 0.00 mol/L. At equilibrium, [SO3]=0.040 mol/L.

What are the Kc and the Kp values?

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%

altitude, density, pressure

air gets less dense the higher the altitude and also the air pressure is less as altitude rises.

a is the answer to this question

explanation:

answer: mole of the gas

answer: mole of the gasexplanation: