Determine K for the following reaction: 1/2 N_2 (g) + 1/2 O_2 (g) + 1/2 Br_2 (g)

Question

Determine K for the following reaction: 1/2 N_2 (g) + 1/2 O_2 (g) + 1/2 Br_2 (g) doubleheadarrow NOBr (g) Given: 2 NO (g) doubleheadarrow N_2 (g) + O_2 (g) K = 2.4 Times 10^30 NO(g) + 1/2 Br_2 (g) doubleheadarrow NOBr(g) K = 1.4 Calculate the pH and percent dissociation for 0.020 M monochloroacetic acid. Write net ionic equations to show what happens when aqueous HCl is added to aqueous potassium nitrite aqueous sodium hydroxide is added to aqueous methylamine hydrochloride aqueous sodium hydroxide is added to aqueous nitrous acid For each reaction in problem 7 (a, b & c), Classify each reactant as an acid or a base; Calculate K; Specify whether it favors reactants, products or neither. Calculate the pH of the following salt solutions: 0.20 M KCN 0.50 M methylamine hydrobromide 0.20 M Li_2 CO_3

Explanation / Answer

5] 1/2N2 + 1/2 O2 + 1/2 Br2 ------> NOBr

Given sequence reactions ;

2NO -----> N2 + O2 ----------------> K1

NO + 1/2 Br2 ------> NOBr ------------> K2

K for the actual reaction is

K = K2 / square root of K1

this is the equilibrium constant .

6] Given 0.020 M

CH2ClCOOH ------------> CH2ClCOO- + H+

0.02 - x x x

Ka = x^2 / 0.02-x

Ka of CH2ClCOOH = 1.3*10^-3

find x value keeping Ka in the above equation ;

x = 4.5*10^-3

pH = -logx = 2.34

percent of dissociation = x *100 / 0.02 = 22.5

7] a] You can only have a net ionic equation where the product(s) is either (g) (s) or (l) . The equation cannot include a product that is (aq) because that indicates that the product is in solution and has dissociated.

So there is no net ionic equation

b] OH- + CH3NH2 ------> H2O + CH3NH-

c] HNO2(aq) + OH-(aq) --> NO2-(aq) + H2O(l)

8] a] Acid and salt

b] Base and base

c] base and acid

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