Author: P Lawson
Last saved by: Peggy Lawson

Name:  ANSWER KEY

Chemistry 30
Unit 3: Chemical Equilibrium
Assignment 2: 2-1 to 2-2

1.	a)	What is meant by a reversible reaction?	

				a reaction that can go in both forward and reverse directions.

	b)	Are all chemical reactions reversible?	No

	c)	Are all reversible reactions always at equilibrium?		No

	d)	Does a reaction have to be reversible in order to reach equilibrium?	Yes

	e)	What, exactly, is equal at equilibrium? (define equilibrium)

					When the rate of the forward reaction equals the rate of the reverse direction.

	f)		How is equilibrium different from a steady state system?

			A steady state occurs in an open system, where there is continual addition of reactants and removal of products. An example would be an assembly line. An equilibrium occurs in a closed system, with no additional reactants being added or products being removed.

2.	Write the equilibrium expression for each of the following reactions. Be sure to pay attention to physical states:

	a)  Br_2 (g) + 5 F_2 (g)  2 BrF_5(g)				

	b)  4 HCl (g) + O_2(g)  2 H_2O (g) + 2 Cl_2 (g)		

	c)  5 Fe^+2(aq) + MnO_4^-(aq) + 8 H^+(aq)  5 Fe^+3(aq) + Mn^+2(aq) + 4 H_2O(l)

3.	For each of the following reactions, state whether the value of the equilibrium constant favours the formation of reactants, products, or both sides equally.

	a)  I_2(g) + Cl_2(g)   2 ICl(g)		K_eq = 2 x 10^6 		^products

	b)  H_2(g) + Cl_2(g)  2 HCl(g)		K_eq = 1.08		equal

	c)  I_2(g)  I(g) + I(g)			K_eq = 3.8 x 10^-7	reactants

4.	Molecular chlorine decomposes into atoms according to the reaction:

		Cl_2 (g)  2 Cl (g)

	The equilibrium constant for the reaction at 25C is 1.4 x 10^-38.  Would many chlorine atoms be present at this temperature?  How do you know?

			The reactants are clearly favored since K_eq is much less than 1. The reactant side of the equation shows molecular chlorine.  Molecular chlorine is thus far more abundant than individual chlorine atoms.

5.	Calculate K_eq for each of the following. Be sure to set up the equilibrium constant expression first, before substituting in the values.  

	Show your work! Pay attention to exponents!

	a)	H_2(g) + Cl_2(g) 2 HCl				[H_2] = 1.0  10^-2 M
									[Cl_2] = 1.0  10^-2 M
									[HCl] = 1.0  10^-2 M

	b)	N_2(g) + 3 H_2(g)  2 NH_3(g)				[N_2] = 4.4  10^-2 M
									[H_2] = 1.2  10^-1 M
									[NH_3] = 3.4  10^-3 M

		

	c)	2 CO(g) + O_2 (g)  2 CO_2 (g)			[CO] = 2.5  10^-3 M
									[O_2] = 1.6  10^-3 M
									[CO_2] = 3.2  10^-2 M

		

	d)	CH_4(g) + H_2O(g)  CO(g) + 3 H_2(g) 			[CH_4] = 2.97  10^-3 M 
									[H_2O] = 7.94  10^-3 M 
									[CO] = 5.45  10^-3 M
									[H_2] = 2.1  10^-3 M 

		

6.	For the following reaction at equilibrium at 2000C, the concentration of N_2 and O_2 are  both 5.2 M.

N_2(g) + O_2(g)  2 NO(g)	K_eq = 6.2  10^-4

	Calculate the concentration of NO at equilibrium. Show your work; pay careful attention to exponents.

			Let x represent the value of the unknown concentration, [NO]

	Substitute the known and unknown values into the equation & solve

				Rearrange to isolate x^2:	

7.	Acetic acid, HC_2H_3O_2, is in equilibrium with its ions:

HC_2H_3O_2(aq)  H^+(aq) + C_2H_3O_2^-(aq)	K_eq = 1.8  10^-5

	At equilibrium, the concentration of the ions are:		[H^+] = 1.33  10^-3 M
										[C_2H_3O_2^-] = 1.33  10^-3 M

	Calculate the concentration of the acid, HC_2H_3O_2.

			  	Let x = the unknown, [HC_2H_3O_2]

			

		Rearrange to isolate x:

Unit 3: Equilibrium	Assignment 2