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3 years ago

What factors could be used to increase reaction rate? Check all that apply.

Chemistry

2 answers:

faust18 [17]3 years ago

6 1

The correct answer is Increasing temperature

ludmilkaskok [199]3 years ago

4 0

The rate of a reaction can be increased by:

-using a catalyst- it lowers the activation energy and leads to less energy required to break bonds of reactants. This lower activation energy leads to more products formed with less time, hence an increase reaction rate.

-increasing temperature- the particles of a molecule move faster and undergo more collision which leads to an increase in the speed of the reaction.

-increasing pressure- means there is more particles of reactants in a reduced volume. The particles do not need to move long distances to find another particle to react with, hence the rate of reaction increases.

Using an inhibitor will not increase the rate of the reaction because, an inhibitor binds to the active site where a catalyst is supposed to act. This means a higher activation energy and thus a decrease in reaction rate.

Likewise, decreasing the concentration implies few particles available to collide with each other and that slow down the speed of the reaction.

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

When a skin diver is on the top of the water at 1.03 atm, her lung volume is 3.62L. How will the volume of her lungs change as s

Anon25 [30]

Answer:

–2.23 L

Explanation:

We'll begin by calculating the final volume. This can be obtained as follow:

Initial pressure (P₁) = 1.03 atm

Initial volume (V₁) = 3.62 L

Final pressure (P₂) = 2.68 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

1.03 × 3.62 = 2.68 × V₂

3.7286 = 2.68 × V₂

Divide both side by 2.68

V₂ = 3.7286 / 2.68

V₂ = 1.39 L

Finally, we shall determine the change in volume. This can be obtained as follow:

Initial volume (V₁) = 3.62 L

Final volume (V₂) = 1.39 L

Change in volume (ΔV) =?

ΔV = V₂ – V₁

ΔV = 1.39 – 3.62

ΔV = –2.23 L

Thus, the change in the volume of her lung is –2.23 L.

NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!

3 0

3 years ago

In a mixture of 2 ideal gases, A and B, PA = 0.2PB, what is the mole fraction of A?

Alborosie

Answer:

0.1667

Explanation:

Hello,

Dalton's law defines:

$y_i=\frac{P_i}{P_T}$

A total pressure is:

$P_T=P_A+P_B$

So, for A (solving for $P_A$ in the previous equation, we get:

$y_A=\frac{P_A}{P_A+P_B}$

Since $P_A=0.2P_B$ , we obtain:

$y_A=\frac{P_A}{P_A+P_A/0.2}\\y_A=\frac{1}{6}\\y_A=0.1667$ }

Best regards.

7 0

3 years ago

How does changing the surface area increase the reaction rate of a chemical change

Anon25 [30]

Effect of increasing surface area on the rate of a reaction. ... Increasing the surface area of a solid reactant exposes more of its particles to attack. This results in an increased chance of collisions between reactant particles, so there are more collisions in any given time and the rate of reaction increases.

6 0

3 years ago

I cant remember my mass, grams, and volume. and how to divide density

Bas_tet [7]

Mass divide by volume

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3 years ago