How does temperature affect the copper (II) chloride equilibrium? Is the forward reaction (color changing from blue to green) endothermic or exothermic? Justify your choice with experimental evidence i. e color changes in the video for Part B.

See explanation

Explanation:

A popular experiment that describes the effect of heat on the position of equilibrum is the change of colour when copper II chloride is heated.

As the solution is heated, it's colour changes from blue to green, this implies the the colour change (blue to green) is an endothermic process (equilibrum position shifts to the right with increase in temperature)

The equilibrum is represented by the equation;

[Cu(H2O)6]^2+(aq) + 4Cl^-(aq)<>[CuCl4]^2-(aq) + 6H2O(l) ∆H=positive

explanation:

crosscutting concepts have application across all domains of science. as such, they are a way of linking the different domains of science. they include patterns; cause and effect; scale, proportion, and quantity; systems and system models; energy and matter; structure and function; and stability and change.

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