The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (M) Initial Rate (10–3 M/s) 0.4 0.4 0.2 160 0.2 0.4 0.4 80 0.6 0.1 0.2 15 0.2 0.1 0.2 5 0.2 0.2 0.4 20 Using the initial-rate method, what is the order of the reaction with respect to C? a. zero-order b. first-order c. third-order d. second-order e. impossible to tell from the data given

Answers

Answer 1

Answer:

The dependence of the power of the reaction rate on the concentration is called the order of the reaction. The order of the reaction is the first order.

What is the initial-rate method?

The initial rate method is the estimation of the order of the reaction by the initial rates of the reactants and products and by performing the reaction several times by measuring the rate.

The reaction is given as,

$\rm A + B + C \rightarrow Products$

The rate of reaction can be given as:

$\rm rate = k[A]^(x)[B]^(y)[C]^(z)$

Here the variables x, y and z are orders respective to the reactant concentration and k is the rate constant.

Value of x with respect to A:

$\begin{aligned} \rm \frac {Rate 3}{Rate 4} &= \rm [([A(3)])/([A(4)])]^(\rm x)\n\n(15)/(5) &= [([0.6])/([0.2])]^(\rm x)\n\n\rm x &= 1\end{aligned}$

Value of y with respect to B:

$\begin{aligned}\rm \frac {Rate 2}{Rate 5} &= \rm [([B(2)])/([B(5)])]^(\rm y)\n\n(80)/(20) &= [([0.4])/([0.2])]^(\rm y)\n\n\rm y &= 2\end{aligned}$

Value of z with respect to C:

$\rm \frac {Rate 1}{Rate 2} &= [([A(1)])/([A(2)])]^(x) [([B(1)])/([B(2)])]^(y) [([C(1)])/([C(2)])]^(z)$

Substituting value of x = 1 and y = 2 in the above equation:

$\begin{aligned}(160)/(80) &= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2) [([0.2])/([0.4])]^(\rm z)\n\n1 &= (0.5)^(\rm z)\n\n&= 1\end{aligned}$

Therefore option b. with respect to C = 1, the order of the reaction is first-order.

Learn more about the order of reaction here:

brainly.com/question/8139015

Answer 2

Answer:

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)

1. 0.4 0.4 0.2 160

2. 0.2 0.4 0.4 80

3. 0.6 0.1 0.2 15

4. 0.2 0.1 0.2 5

5. 0.2 0.2 0.4 20

The rate of the above reaction is given as:

$Rate = k[A]^(x)[B]^(y)[C]^(z)$

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

$(Rate3)/(Rate4)= [([A(3)])/([A(4)])]^(x)$

$(15)/(5)= [([0.6])/([0.2])]^(x)$

$3 = 3^(x) \n\nx =1$

Order w.r.t B : Use trials 2 and 5

$(Rate2)/(Rate5)= [([B(2)])/([B(5)])]^(y)$

$(80)/(20)= [([0.4])/([0.2])]^(y)$

$4 = 2^(y) \n\ny =2$

Order w.r.t C : Use trials 1 and 2

$(Rate1)/(Rate2)= [([A(1)])/([A(2)])]^(x)[([B(1)])/([B(2)])]^(y)[([C(1)])/([C(2)])]^(z)$

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

$(160)/(80)= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2)[([0.2])/([0.4])]^(z)$

$1 = (0.5)^(z)$

z = 1

Therefore, order w.r.t C = 1

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During an experiment, a student adds 2.90 g CaO to 400.0 mL of 1.500 M HCl . The student observes a temperature increase of 6.00 °C . Assuming that the solution's final volume is 400.0 mL , the density is 1.00 g/mL , and the heat capacity is 4.184 J/g⋅°C , calculate the heat of the reaction, ΔHrxn .

Answers

Answer:

ΔHrxn = 193107.69 J/mol

Explanation:

ΔHrxn = mcΔT

m = mass

c = heat capacity

ΔT = temperature variation

density = m/V

m = density x V

m = 1.00 g/mL x 400.0 mL

m = 400.0 g

ΔHrxn = mcΔT

ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C

ΔHrxn = 10041.6 J

CaO + 2HCl → CaCl₂ + H₂O

CaO = 56.0774 g/mol

2.90 g CaO = 0.052 mol

400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl

ΔHrxn = 10041.6 J is for 0.052 mol of CaO

ΔHrxn = 193107.69 J is for 1 mol of CaO

Calculate the concentration of OH in a solution that contains 3910-4 M H30 at 25°C. Identify the solution as acidic, basic or neutral OA) 2.6 10-11 M, acidic OB)26 10-11 M. basic O c) 3.9 x 10-4 M, neutral OD) 2.7 * 10-2 M

Answers

Answer : The correct option is, (A) $2.6* 10^(-11)M$ , acidic

Explanation:

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

When the value of pH is less then 7 then the solution will be acidic.

When the value of pH is more then 7 then the solution will be basic.

When the value of pH is equal to 7 then the solution will be neutral.

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (3.9* 10^(-4))$

$pH=3.41$

Now we have to calculate the pOH.

$pH+pOH=14\n\npOH=14-pH\n\npOH=14-3.41=10.59$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10.59=-\log [OH^-]$

$[OH^-]=2.6* 10^(-11)M$

Therefore, the $OH^-$ concentration is, $2.6* 10^(-11)M$

Calculate the individual percent recoveries of benzoic acid, naphthalene and 3-nitroaniline if you were able to collect 9.75 g of benzoic acid, 6.41 g of naphthalene, and 7.71 g of 3-nitroaniline from a set of extractions. The starting mass of the mixture was 26.24 g. (0.6 pt)

Answers

Answer:

Benzoic acid= 37.16%

Naphthalene = 24.43%

3-Nitroaniline= 29.38%

Explanation:

Data given:

percentage recovery of benzonic acid = 9.75/26.24 * 100 = 37.16%

Percentage recovery of napthalene = 6.41/26.24 * 100 = 24.43%

Percentage recovery of 3-nitroaniline = 7.71/26.24 * 100 = 29.38%

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K. A sample of C8H18 is placed in a closed, evacuated 537 mL container at a temperature of 339 K. It is found that all of the C8H18 is in the vapor phase and that the pressure is 68.0 mm Hg. If the volume of the container is reduced to 338 mL at constant temperature, which of the following statements are correct?a. No condensation will occur.
b. Some of the vapor initially present will condense.
c. The pressure in the container will be 100. mm Hg.
d. Only octane vapor will be present.
e. Liquid octane will be present.

Answers

Answer:

the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Explanation:

Given that;

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K

Initial volume of the container, V1 = 537 mL

Initial vapor pressure, P1 = 68.0 mmHg

Final volume of the container, V2 = 338 mL

Let us say that the final vapor pressure = P2

From Boyle's law,

P2V2 = P1V1

P2 * 338 = 68.0 * 537

338P2 = 36516

P2 = 36516 / 338

P2 = 108.03 mmHg

Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Suppose a scientist made a claim that all spontaneous reactions are exothermic. Whic of the following would provide the strongest challenge to their claim? Suppose a scientist made a claim that all spontaneous reactions are exothermic. Which of the following would provide the strongest challenge to their claim? a. An exothermic reaction which is not spontaneous
b. An endothermic reaction that only proceeds when coupled to an exothermic reaction
c. An endothermic reaction that only proceeds when a catalytst is present
d. An endothermic reaction which is not spontaneous
e. All of the above

Answers

Answer: Option (c) is the correct answer.

Explanation:

It is given that the scientist is claiming that all the spontaneous reactions are exothermic in nature.

And, it is known that when a reaction is spontaneous in nature then $\Delta G$ is negative.

Now, the relation between Gibb's free energy, enthalpy and entropy is as follows.

$\Delta G$ = $\Delta H - T \Delta S$

So, when a catalyst is present in a chemical reaction then we do not need to give large amount of heat from outside. And, because of this the enthalpy of reaction will not be highly positive.

Hence, the value of $\Delta G$ will result in a negative value which means the reaction is spontaneous.

Thus, we can conclude that an endothermic reaction that only proceeds when a catalytst is present, would provide the strongest challenge to their claim.

1.6x10^23 lead atoms. Find the weight in grams

Answers

Moles of lead(Pb) = 1.6x10^23/6.02x10^23 = 0.265 moles.

Weight of lead = moles x atomic weight of lead
= 0.265x207.2
= 54.908 grams.

Hope this helps!

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