Be sure to answer all parts. write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 δ[ch4] δt = − 1 2 δ[h2o] δt = − δ[co2] δt = 1 4 δ[ch3oh] δt (click in the answer box to activate the palette. do not include states of matter.) b) rate = − 1 2 δ[n2o5] δt = 1 2 δ[n2] δt = 1 5 δ[o2] δt (click in the answer box to activate the palette. do not include states of matter.) c) rate = − 1 2 δ[h2] δt = − 1 2 δ[co2] δt = − δ[o2] δt = 1 2 δ[h2co3] δt (click in the answer box to activate the palette. do not include states of matter.)

Answer : The balanced equations will be:

(a)

(b)

(c)

Explanation :

The general rate of reaction is,

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a)

The balanced equations will be:

(b)

The balanced equations will be:

(c)

The balanced equations will be:

where is the following

explanation:

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