Ask question

Ask question

All categories

3 years ago

11

Be sure to answer all parts. Determine the overall orders of the reactions to which the following rate laws apply: (a) rate = k[

NO2]2 (b) rate = k zero order first order 1.5 order second order 2.5 order third order zero order first order 1.5 order second order 2.5 order third order (c) rate = k[H2][Br2]1/2 (d) rate = k[NO2]2[O2] zero order first order 1.5 order second order 2.5 order third order zero order first order 1.5 order second order 2.5 order third order

1 answer:

noname [10]3 years ago

6 0

Answer:

(a). second order.

(b). zeroth order.

(c). one and a half order.

(d). third order.

Explanation:

Hello,

In this case, you just have to add the concentration's powers in each rate law:

(a) $r=k[NO_2]^2$

Just the 2-power is present, so it is a second order law.

(b) $r=k$

No power is present, so it is a zeroth order law.

(c) $r=k[H_2]*[Br_2]^{1/2}$

1 plus 1/2 gives 1.5, that is why it is a 1.5 (one and a half) order law.

(d) $r=k[NO_2]^2[O_2]$

2 plus 1 gives 3, that is why it is a third order law.

Best regards.

You might be interested in

Indicate what happens to the concentration of Pb2+ in each half cell

Taya2010 [7]

Answer:

Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below

The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.

Explanation:

The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.

In the Left Beaker (Left half cell), their is less concentration

Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side

In the Right Beaker (right half cell), their is more concentration

Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side

Electrons will flow from Left to Right direction.

3 0

3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

<u>We have:</u>

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

2 years ago

I’LL MAKE YOU BRAINLIEST+ FREE POINTS

Pavel [41]

Answer:

The solution's new volume is 1.68 L

Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

Ci*Vi=Cf*Vf

where C and V are concentration and volume, respectively; and the i and f subscripts indicate initial and final respectively.

In this case, being:

Ci= 7 M
Vi= 0.60 L
Cf= 2.5 M
Vf=?

Replacing:

7 M*0.60 L= 2.5 M* Vf

Solving:

$Vf=\frac{7 M*0.60 L}{2.5 M}$

Vf= 1.68 L

<u><em>The solution's new volume is 1.68 L</em></u>

4 0

3 years ago

Do free electrons appear anywhere in the balanced equation for a redox reaction?

Andru [333]

Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.

Explanation:

A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).

The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).

Here, the copper ions gain electrons and become solid copper.

4 0

3 years ago

Read 2 more answers

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

3 0

2 years ago