Be sure to answer all parts. enter your answers in scientific notation. the following reactions have the indicated equilibrium constants at a particular temperature:
n2(g) + o2(g) ⇌ 2no(g) kc = 4.3 × 10−25 2
no(g) + o2(g) ⇌ 2no2(g) kc = 6.4 × 109

determine the values of the equilibrium constants for the following equations at the same temperature:
(a) 4no(g) ⇌ n2(g) + 2no2(g) × 10
(b) 4no2(g) ⇌ 2n2(g) + 4o2(g) × 10
(c) 2no(g) + 2no2(g) ⇌ 3o2(g) + 2n2(g) × 10

a) 1,5x10³⁴

b) 1,5x10¹⁵

c) 7,3x10¹⁴

Explanation:

It is possible to sum different reactions with its equilibrium constant to obtain the equilibrium constant of the new reaction, thus:

a) 2 NO(g) ⇄ N₂(g) + O₂(g) kc = 1/4,3x10⁻²⁵

+ 2 NO(g) + O₂(g) ⇄ 2 NO₂(g) kc = 6,4x10⁹

= 4 NO(g) ⇄ N₂(g) + 2NO₂(g) Where its equilibirum constant is:

1/4,3x10⁻²⁵ × 6,4x10⁹ = 1,5x10³⁴

b) 4 NO(g) ⇄ 2 N₂(g) + 2 O₂(g) kc = 2× 1/4,3x10⁻²⁵

4 NO₂(g)  ⇄ 4 NO(g) + 2 O₂(g) kc = 2× 1/6,4x10⁹

= 4 NO₂(g) ⇄ 2 N₂(g) + 4 O₂(g) Where its equilibirum constant is:

2× 1/4,3x10⁻²⁵ × 2× 1/6,4x10⁹ = 1,5x10¹⁵

c) 4 NO(g) ⇄ 2 N₂(g) + 2 O₂(g) kc = 2× 1/4,3x10⁻²⁵

2 NO₂(g)  ⇄ 2 NO(g) + O₂(g) kc =  1/6,4x10⁹

= 2NO(g) + 2 NO₂(g) ⇄ 2 N₂(g) + 3 O₂(g) Where its equilibirum constant is:

2× 1/4,3x10⁻²⁵ × 1/6,4x10⁹ = 7,3x10¹⁴

I hope it helps!

the answer is:  

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).

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