Answered

Discover how IDNLearn.com can help you learn and grow with its extensive Q&A platform. Our platform is designed to provide reliable and thorough answers to all your questions, no matter the topic.

1.25mole Nocl was placed in a2.50l reaction chamber at 427c. After equilibrium was reached .1.10moles of Nocl remained.Calculate the equilibrium constant Kc for the reaction. 2Nocl(g)=2No(g)+cl2(g)

Sagot :

Anfabba15idn

Answer:

Kc → 5.58×10⁻⁴

Explanation:

Equilibrium reaction is:

2NOCl (g) ⇄  2NO (g) + Cl₂(g)

Initially we have 1.25 moles of NOCl

After the equilibrium, we have 1.10 moles. So, during the process:

(1.25 mol - 1.1 mol) = 0.15 moles have reacted.

As ratio are 2:2, and 2:1, 0.15 moles of NO and (0.15 /2) = 0.075 moles of chlorine, were produced in the equilibrium.

Finally in equilibrium we have: 1.10 moles of NOCl, 0.15 moles of NO and 0.075 moles of Cl₂. But these amount are not molar, so we need molar concentration in order to determine Kc:

1.10 mol /2.50L = 0.44 M

0.15 mol / /2.50L = 0.06 M

0.075 mol /2.50L = 0.03 M

Let's make expression for Kc → [Cl₂] . [NO]² / [NOCl]²

Kc = (0.03 . 0.06²) / 0.44² → 5.58×10⁻⁴

We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is dedicated to providing accurate answers. Thank you for visiting, and see you next time for more solutions.