Question

Number 2 is confusing a little bit.

1. (5 pts) Write the equilibrium constant K expression for the reaction below. C(s) + 2 H2 (9) = CH(9) Ke=[CH] [H₂] 2. (10 pts) In the reaction below, the reaction flask initially contained 0.50 M NO and 0.50 M Cl2. After the reaction reached equilibrium the concentration of NOCI was measured to be 0.25 M. What is the equilibrium constant Kc? Show all your work + 2 NOCI (9) 2 NOC49) 2 NOg I 0.50 -0.25 0.25 2 NO (9) + Clz (g) - Cl29 ² 0.50m -0. 25 0. 25 +0. 25 0.25 Kc = [NOCL]². [NO]" [CL2]

Answer 1

2-

The given chemical reaction is- 2NO + Cl2 --------> 2NOCl

Now equilibrium constant (Kc) for a reaction is the ratio between the product of concentration of products, each raised to the power of number of moles involved to the product of concentration of reactants, each raised to the power of number of moles involved, at equilibrium.

i.e for the given reaction-

Kc = [NOCl]2 / [NO]2 * [Cl2] where all the concentrations are at the time of equilibrium.

Now putting the given values in the ICE table

Reaction	2NO +	Cl2 -------->	2NOCl
Initial	0.50 M	0.50 M	0
Change	- 2*0.125‬	-0.125‬	+ 2 * 0.125‬
Equilibrium	0.25‬ M	0.375‬ M	0.25 M

So putting the vlaues-

Kc = [NOCl]2 / [NO]2 * [Cl2]

= [0.25]2 / [0.25]2 * [0.375]

= 2.66

when 2 M of NOCl is formed, the decrease form NO should be 2M where as for Cl2 it is 1 M