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Dmitriy789 [7]
4 years ago
9

A chemical engineer is developing a process for producing a new chemical. One step in the process involves allowing a solution o

f potassium hydroxide to react with a solid. Which action would most likely increase the reaction rate for this step?
Chemistry
2 answers:
notka56 [123]4 years ago
8 0

increasing temperature

alexgriva [62]4 years ago
5 0
<h2>Answer with Explanation </h2>

An action of enhancing the temperature would be most feasible to improve the rate of reaction as the kinetic energy increments which generates the particles to collide with each other quicker. The rate of reaction duplicates for every 10 degree Celsius rise in temperature which causes the particles to have more collisions vigorously at an intense rate. The frequency of collisions also increases giving more speed to the rate of reaction which reaches energies equal to or greater than activation energy.

You might be interested in
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
4 years ago
0.467 mol NaCl are needed to
Alisiya [41]

Answer:

27.4 gram is the solution it's simple dude...

Explanation:

don't be afraid of huge question they confuse you you need not to be confused

now see simple solution

molality is denoted by m

so

m= moles of solute / mass of solvent in kg.

i hope your know the meaning of solute and solvent....

so moles are given 0.467

and molar mass is given 58.44

so just take out the gram means

by applying formula

58.44×0.467

it will give 27.4 grams simple.....

8 0
2 years ago
Your tongue and nose work separately.<br> True<br> False
yuradex [85]
False, our tongue and nose work together
3 0
3 years ago
Read 2 more answers
If one adds 0.1 mol of the weak acid hf (pk_a = 3.2) to a solution with a ph = 2, which species would be most abundant and how m
zavuch27 [327]

F^{-} is most abundant and 6310 times more than HF.

<h3>What is a strong and weak acid?</h3>

When an acid is dissolved in water, all of its molecules disintegrate, making the acid powerful.

When an acid is dissolved in water, only a small number of its molecules disintegrate, making the acid weak. Strong acids have a lower pH than weak acids.

The powerful acids include perchloric acid, chloric acid, nitric acid, sulfuric acid, hydrobromic acid, and hydroiodic acid.

Given:

Pka=3..2

pH=7

Let the volume be 1 liter

[HF]=01 M

pH=pka+log \frac{F^{-}}{HF} \\\\7=3.2+log\frac{F^{-}}{HF} \\3.8=log\frac{F^{-}}{HF}\\ \frac{F^{-}}{0.1}=10^{3.8} \\F^{-}=630.95 M

Now,

\frac{F^{-}}{HF}=\frac{630.95}{0.1}\\ =6309.57

F-:HF= 6309.57:1

Therefore, the most abundant is F^{-}and has 6310 times more than HF is F^{-}.

To know more about strong and weak acids, visit: brainly.com/question/12811944

#SPJ4

4 0
2 years ago
What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i
Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314  (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0
3 years ago
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