Answer:
The answer to your question is [NObr] = 1.41
Explanation:
Data
Kc = 2
[NO] = 2 M
Br₂ = 1.0 M
NOBr = ?
Process
1.- Write the Equation of the Kc
Kc = [tex]\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}}[/tex]
2.- Solve for [NOBr]
[NOBr]² = [tex]\frac{[NO]^{2}[Br_{2}]}{Kc}[/tex]
3.- Substitution
[NOBr]² = [tex]\frac{[2]^{2}[1]}{2}[/tex]
4.- Simplification
[¨NOBr]² = [tex]\frac{4}{2}[/tex]
[NOBr]² = 2
5.- Result
[NOBr] = [tex]\sqrt{2}[/tex]
[NOBr] = 1.41
Answer:
The molar concentration of NOBr is 1.414 M
Explanation:
Step 1: Data given
Kc = 2.0
The equilibrium mixture contains:
2.0 M NO
1.0 M Br2
Step 2: The balanced equation
2NOBr (g) ⇌ 2NO(g) + Br2 (g) Kc = 2.0
Step 3: Define Kc
Kc = [Br2]*[NO]²/[NOBr]²
⇒ with Kc = 2.0
⇒with [Br2] = 1.0M
⇒with [NO] = 2.0 M
⇒with [NOBr] = TO BE DETERMINED
Step 4: Calculate the molar concentration of NOBr
Kc = [Br2]*[NO]²/[NOBr]²
2.0 = (1.0*2.0²) / NOBr²
NOBr² = 4.0 / 2.0 = 2.0
NOBr = √2.0 = 1.414 M
The molar concentration of NOBr is 1.414 M
