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3 years ago

What is the rate of a reaction if the value of k is 3 and A and B are each 2 M Rate kA2B?

Chemistry

1 answer:

4vir4ik [10]3 years ago

7 0

Answer:

Rate= 24

Explanation:

Rate = k[A]^2[B]

k= 3, [A]=2M, [B]=2M

Rate= 3×(2)^2×2= 24

Substitute into above

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2. A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine:2HI (g) H2(g)

Deffense [45]

Answer:

The value of Kp at this temperature is 7.44*10⁻³

Explanation:

Chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.

For the general chemical equation for a homogeneous gas phase system:

aA + bB ⇔ cC + dD

where a, b, c and d are the stoichiometric coefficients of compounds A, B, C and D, the equilibrium constant Kp is determined by the following expression:

$Kp=\frac{P_{C} ^{c} *P_{D} ^{d} }{P_{A} ^{a} *P_{B} ^{b} }$

Where Px is the partial pressure of each of the components once equilibrium has been reached and they are expressed in atmospheres. The equilibrium constant Kp depends solely on temperature and is dimensionless.

In the case of the reaction:

2 HI (g) ⇔ H₂ (g) + I₂ (g)

the equilibrium constant Kp is determined by the following expression:

$Kp=\frac{P_{H_{2} } *P_{I_{2} } }{P_{HI} ^{2} }$

The system comes to equilibrium at 425 °C, and

PHI = 0.794 atm
PH2 = 0.0685 atm
PI2 = 0.0685 atm

Replacing:

$Kp=\frac{0.0685*0.0685}{0.794^{2} }$

Kp=7.44*10⁻³

<u><em>The value of Kp at this temperature is 7.44*10⁻³</em></u>

4 0

3 years ago

identify 2 changes of state that require energy to be used, and 2 changes of state that require the release of energy

solniwko [45]

Answer:

used=Evaporation and sublimation

released=condensation and freezing

Explanation:

8 0

3 years ago

An acid is:

agasfer [191]

An acid is a proton donor

5 0

3 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

What role does valence electrons play in reactivity for an element?

kirza4 [7]

Explanation:

The valence electrons within an atom is the number of electrons in its outermost shell.

These electrons are used by an atom to react with one another. They determine the extent to which an atom is ready to combine either by losing, gaining or sharing these electrons.

Every atom desires to have a completely filled outermost shell.
Only the elements in group 8 have a complete octet.
The need to attain stability is driven by the number of electrons in their valence shell.
Therefore, some atoms are very reactive.
Those needing one electrons to complete their octet and also those that must lose one electron are very reactive.

4 0

2 years ago