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BabaBlast [244]
3 years ago
11

Consider the reaction between no and cl2 to form nocl. 2no(g) + cl2(g) ⇌ 2nocl(g) a reaction mixture at a certain temperature in

itially contains only [no] = 0.50 m and [cl2] = 0.50 m. after the reaction comes to equilibrium, the concentration of nocl is 0.30 m. what is the value of kc at this temperature?
Chemistry
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

1.125 x 10⁻³.

Explanation:

  • For the reaction: <em>2NO(g) + Cl₂(g) → 2NOCl(g).</em>

<em>Kc = [NOCl]²/[NO]²[Cl₂],</em>

Where, [NOCl], [NO], and [Cl₂] is the equilibrium concentration of NOCl, NO, and Cl₂.

∵ Initial concentration of NO and Cl₂ = 0.50 M.

<em>And the equilibrium concentration of NOCl = 0.30 M.</em>

<em />

∴ The equilibrium concentration of NO, and Cl₂: [NO] = [Cl₂] = 0.50 M - 0.30 M = 0.20 M.

<em>∴ Kc = [NOCl]²/[NO]²[Cl₂]</em> = [0.30]²/[0.20]²[0.20] = <em>1.125 x 10⁻³.</em>

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A buffer solution contains 0.240 M ammonium chloride and 0.499 M ammonia. If 0.0565 moles of perchloric acid are added to 250 mL
snow_lady [41]

Answer:

The pH of the resulting solution is 9.02.

Explanation:

The initial pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.240}{0.499}) = 4.43        

pH = 14 - pOH = 14 - 4.43 = 9.57

Now, the perchloric acid added will react with ammonia:

n_{NH_{3}} = 0.499moles/L*0.250 L - 0.0565 moles = 0.0683 moles

Also, the moles of ammonium chloride will increase in the same quantity according to the following reaction:

NH₃ + H₃O⁺ ⇄ NH₄⁺ + H₂O    

n_{NH_{4}Cl} = 0.240 moles/L*0.250L + 0.0565 moles = 0.1165 moles

Finally, we can calculate the pH of the resulting solution:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.1165 moles/0.250 L}{0.0683 moles/0.250 L}) = 4.98  

pH = 14 - 4.98 = 9.02

Therefore, the pH of the resulting solution is 9.02.

I hope it helps you!

5 0
3 years ago
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