Consider the following system at equilibrium at 573 k: 2nocl(g) + 18.4 kcal2no(g) + cl2(g) if the temperature on the equilibrium system is suddenly increased: the reaction must: a. run in the forward direction to restablish equilibrium. b. run in the reverse direction to restablish equilibrium. c. remain the same. already at equilibrium. the concentration of cl2 will: a. increase. b. decrease. c. remain the same.

After increasing the temperature of the system:

The reaction must: Run in the forward direction to restablish equilibrium.

Already at equilibrium. The concentration of Cl2 will: Increase.

Explanation:

The equilibrium constant, for an endothermic reaction (ΔH ° positive), increases with increasing temperature.

The principle of Le Châtelier indicates that if an equilibrium mixture is heated, and consequently its temperature increases, there will be a net reaction in the direction that counteracts the disturbance caused by the increase in temperature. For an endothermic reaction, the direct reaction absorbs energy in the form of heat.

Consequently, the equilibrium shifts to the side of the products, which means that Kc increases with increasing temperature.

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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