Find the equilibrium constants, kp, for the following equilibria, (i) no(g) + ½ o2(g) ⇄ no2(g), kp = ? (ii) no2(g) ⇄ no(g) + ½ o2(g), kp = ? , given the equilibrium constant, kp, for the reaction: 2no (g) + o2(g) ⇄ 2no2(g) kp= 100 at the same temperature

Answer :

(i) The value of equilibrium constants for this reaction is, 10

(ii) The value of equilibrium constants for this reaction is, 0.1

Explanation :

The given equilibrium reaction is,

Now we have to determine the equilibrium constants for the following equilibrium reactions.

(i)

From the given reaction we conclude that, the reaction (i) will takes place when the given main reaction will be multiplied by half (1/2). That means when reaction will be half then the equilibrium constant will be:

The value of equilibrium constants for this reaction is, 10

(ii)

From the given reaction we conclude that, the reaction (ii) will takes place when the reaction (i) will be reverse. That means when reaction will be reverse then the equilibrium constant will be:

The value of equilibrium constants for this reaction is, 0.1

pure substance: iron,copper

impure substance: water,charcoal

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years