As an epa scientist studying catalytic converters and urban smog, you want to find kc for the following reaction: 2no2(g) ⇋ n2(g) + 2 o2(g) kc = ? . use the following data to find the unknown kc (1) ½ n2(g) + 1/2 o2(g) ⇋ no(g) kc = 4.8x10-10 (2) 2no2(g) ⇋ 2no(g) + o2(g) kc = 1.1x10-5

Kc = 4.774 * 10¹³

Explanation:

the desired reaction is

2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)

Kc =[N₂]*[O₂]² /[NO₂]²

Since

1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)

Kc₁= [NO]/(√[N₂]√[O₂]) →  Kc₁²= [NO]²/([N₂][O₂])

and

2 NO₂(g) ⇋ 2 NO(g) + O₂(g)

Kc₂= [NO]²*[O₂]/[NO₂]² →  1/Kc₂= [NO₂]²/([NO]²[O₂])

then

Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂])  = [NO₂]²/([N₂]*[O₂]²) = 1/Kc

Kc₁² /Kc₂ = 1/Kc

Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³

what are the statements

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2