Equilibrium Mass action Expression and Equilibrium constant

What is the K_c expression for this reaction?
   2 NO_(g)  +  O_2(g) ⇌  2 NO_2(g)

• K_c = [NO₂]² / [NO]² [O₂]
• K_c =  [NO]² [O₂] / [NO₂]²
• K_c = [NO]² [O₂] [NO₂]²
• K_c = [NO₂]² / [NO]² +  [O₂]

What is the K_c expression for this reaction?
   2 NO_(g)  +  O_2(g) ⇌  2 NO_2(g)

• K_c = [NO₂]² / [NO]² [O₂]
• K_c =  [NO]² [O₂] / [NO₂]²
• K_c = [NO]² [O₂] [NO₂]²
• K_c = [NO₂]² / [NO]² +  [O₂]

What are [A] and [B]?

• concentration of reactants
• concentration of products
• energy of reactants
• energy of products

The following reaction :    
SO_{2 (g)} +  NO_{2 (g)} <=> SO_{3 (g)} + NO (g)  
had reached a state of equilibrium, was found to contain  
0.40 mol L^-1 SO₃ , and 0.30 mol L^-1 NO,
0.15 mol L^-1NO₂ , and 0.20 mol L^-1 SO₂.
Calculate the equilibrium constant  for this reaction. 

• 4
• .42
• .25
• 1

What is the equilibrium expression for:
Fe₃O₄(s) + 4H₂(g) <=> 3Fe(s) + 4H₂O(g)
K_c =

• [Fe]³ [H₂O]⁴  / [Fe₃O₄] [H₂]⁴
• [Fe₃O₄] [H₂]⁴ /  [Fe]³ [H₂O]⁴
• [H₂O]⁴ / [H₂]⁴
• [Fe] [H₂O] / [Fe₃O₄] [H₂]

If the equilibrium constant is much greater than one (K >> 1) then

• Equilibrium is not established.
• Equilibrium lies to the right (products are favored)
• Equilibrium lies to the left (reactants are favored)
• Neither products or reactants are favored.

If the equilibrium constant is much less than one (K << 1) then

• Equilibrium is not established.
• Equilibrium lies to the right (products are favored)
• Equilibrium lies to the left (reactants are favored)
• Neither products or reactants are favored.

What is the equilibrium-constant expression for 
CO₂(g) + H₂(g) ↔ CO(g) + H₂O(l)

• K_c= [CO][H₂O] / [CO₂][H₂]  
• K_c= [CO₂][H₂] / [CO]
• K_c= [CO₂][H₂] / [CO][H₂O]
• K_c= [CO] / [CO₂][H₂]  

N₂O₄(g) ↔ 2 NO₂(g)
What is the concentration equilibrium constant expression?

• K_c = [NO₂]²/[N₂O₄]
• K_c = [N₂O₄]/[NO₂]²
• K_c = [N₂O₄]²/[NO₂]
• K_c = [NO₂]/[N₂O₄]²

Find the K_c for
2NOBr(g) + Cl₂(g) ↔ 2NO(g) + 2BrCl(g)
 if 
2NOBr(g) ↔ 2NO(g) + Br₂(g)
has a K_c = 0.014
Br2(g) + Cl2(g) ↔ 2BrCl(g)
has a K_c = 7.2

• 0.10
• 10
• 1.0 x 10^-3
• 1.0 x 10^-2