Which reaction results in the GREATEST increase in entropy?

. In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in

valentina_108 [34]

Answer : The the rate expression will be:

$Rate=-\frac{d[Cl_2]}{dt}=-\frac{1}{3}\frac{d[F_2]}{dt}=+\frac{1}{2}\frac{d[ClF_3]}{dt}$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

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 :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

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 rate expressions for this given reaction.

The given balanced equations is:

$Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)$

$\text{Rate of disappearance of }Cl_2=-\frac{d[Cl_2]}{dt}$

$\text{Rate of disappearance of }F_2=-\frac{1}{3}\frac{d[F_2]}{dt}$

$\text{Rate of formation of }ClF_3=+\frac{1}{2}\frac{d[ClF_3]}{dt}$

Thus, the rate expression will be:

$Rate=-\frac{d[Cl_2]}{dt}=-\frac{1}{3}\frac{d[F_2]}{dt}=+\frac{1}{2}\frac{d[ClF_3]}{dt}$

5 0

3 years ago

What is an isolated system

ratelena [41]

Answer would be A. Why because isolation means that nothing can move in or out

3 0

3 years ago

Why are some compounds that contain carbon not considered as organic compounds?

Arlecino [84]

Because it may not contain Hydrogen or Oxygen

7 0

3 years ago

If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?

andre [41]

Complete question:

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.44×10⁻⁴ s⁻¹ at a certain temperature.

If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?

Answer:

After 210 s the concentration of SO2Cl2 will be 0.121 M

Explanation:

$ln\frac{[A_t]}{[A_0]} =-kt$

where;

At is the concentration of A at a time t

A₀ is the initial concentration of A

k is rate constant = 1.44×10⁻⁴ s⁻¹

t is time

ln(At/A₀) = -( 1.44×10⁻⁴)t

ln(At/0.125) = -( 1.44×10⁻⁴)210

ln(At/0.125) = -0.03024

$\frac{A_t}{0.125} = e^{-0.03024$

At/0.125 = 0.9702

At = 0.125*0.9702

At = 0.121 M

Therefore, after 210 s the concentration of SO2Cl2 will be 0.121 M

6 0

3 years ago

Consider the interaction of two hydrogen ls atomic orbitals of the same phase. Which statements below is an incorrest descriptio

Andreyy89

Answer:

B) The molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital.

Explanation:

When two atoms of hydrogen come close to each other , there is formation of molecular orbital . Due to overlap of 1 s orbital of one and 1 s orbital of another atom , two molecular orbitals are formed . One of these molecular orbital has energy less than 1 s atomic orbital . It is called 1 s sigma bonding molecular orbital . The other molecular orbital has energy more than 1 s atomic orbital . It is called antibonding molecular orbital . Two electrons occupy bonding sigma molecular orbital .

So , the statement that "the molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital " is wrong .

4 0

3 years ago