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Reaction Rates

inn [45]

Answer:

PART A 1st order in A and 0th order in B

Part B The reaction rate increases

Explanation:

PART A

The rate law of the arbitrary chemical reaction is given by

$-r_A=k\times\left[A\right]^\alpha\times\left[B\right]^\beta\bigm$

Replacing for the data

Expression 1 $0.00639=k\times{0.12}^\alpha\times{0.22}^\beta$

Expression 2 $0.01280=k\times{0.24}^\alpha\times{0.22}^\beta$

Expression 3 $0.00639=k\times{0.12}^\alpha\times{0.11}^\beta$

Making the quotient between the fist two expressions

$\frac{0.00639}{0.01280}=\left(\frac{0.12}{0.24}\right)^\alpha$

Then the expression for $\alpha$

$\alpha=\frac{ln\frac{0.00639}{0.01280}}{ln\frac{0.12}{0.24}}=1\bigm$

Doing the same between the expressions 1 and 3

$\frac{0.00639}{0.00639}=\left(\frac{0.22}{0.11}\right)^\beta$

Then

$\beta=\frac{ln\frac{0.00639}{0.00639}}{ln\frac{0.22}{0.11}}=0\bigm$

This means that the reaction is 1st order respect to A and 0th order respect to B .

PART B

By the molecular kinetics theory, if an increment in the temperature occurs, the molecules will have greater kinetic energy and, consequently, will move faster. Thus, the possibility of colliding with another molecule increases. These collisions are necessary for the reaction. Therefore, an increase in temperature necessarily produces an increase in the reaction rate.

4 0

3 years ago

Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati

ollegr [7]

Explanation:

Let us assume that the given data is as follows.

mass of barium acetate = 2.19 g

volume = 150 ml = 0.150 L (as 1 L = 1000 ml)

concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

$Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}$

Hence, moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times Ba(C_{2}H_{3}O_{2})_{2}$ .......... (1)

As, No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of $Ba(C_{2}H_{3}O_{2})_{2}$ will be calculated as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{2.19 g}{255.415 g/mol}$ (molar mass of $Ba(C_{2}H_{3}O_{2})_{2}$ is 255.415 g/mol)

= $8.57 \times 10^{-3}$

Moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times 8.57 \times 10^{-3}$

= 0.01715 mol

Hence, final molarity will be as follows.

Molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.01715 mol}{0.150 L}$

= 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0

3 years ago

What causes high and low tides on Earth? A. The moon's gravitational pull on Earth causes water to bulge on two sides of Earth.B

sveticcg [70]

Answer:

A.

The moon's gravitational pull on Earth causes water to bulge on two sides of the Earth.

7 0

3 years ago

Arrange the following elements in order from lowest to highest first ionization energy : , Ar , Cs, Na *

Sauron [17]

Lowest Cs, Na, highest is At

6 0

3 years ago

Read 2 more answers

A container is filled with a gas. The container has a wall controlled by a moveable piston. The

ioda

Answer:

The new volume will be 3.67 L.

Explanation:

As the volume increases, the gas particles (atoms or molecules) take longer to reach the walls of the container and therefore collide with them fewer times per unit of time. This means that the pressure will be lower because it represents the frequency of collisions of the gas against the walls. In this way pressure and volume are related, determining Boyle's law which says:

"The volume occupied by a certain gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

P*V=k

Now it is possible to assume that you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:

P1 * V1 = P2 * V2

In this case:

P1= 1.85 atm
V1= 4.64 L
P2= 2.34 atm
V2= ?

Replacing:

1.85 atm* 4.64 L= 2.34 atm* V2

Solving:

$V2= \frac{1.85 atm* 4.64 L}{2.34 atm}$

V2= 3.67 L

The new volume will be 3.67 L.

6 0

3 years ago