Answer to Question #337122 in Chemistry for ash

Question

2NOBr(g) 2NO(g) + Br₂(g) 

If 0.364 moles of NOBr(g), 0.448 moles of NO, and 0.543 moles of Br₂ are at equilibrium in a 19.1 L container at 464 K, the value of the equilibrium constant, K_c, is  .

Answer 1

Solution:

Balanced chemical equation:

2NOBr(g) → 2NO(g) + Br₂(g) 

The expression for the equilibrium constant K_c:

[Br₂] = (0.543 mol) / (19.1 L) = 0.02843 M

[NO] = (0.448 mol) / (19.1 L) = 0.02346 M

[NOBr] = (0.364 mol) / (19.1 L) = 0.01906 M

Thus,

K_c = (0.02843) × (0.02346)² / (0.01906)² = 0.043

K_c = 0.043 M

Answer: The value of the equilibrium constant (K_c) is 0.043