1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
siniylev [52]
2 years ago
9

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

50 mol dm^-3 in B. After 1.0 h the concentration of A had fallen to 0.020 mol dm^-3. a) Calculate the Rate constant. b) Solve for the half- life of each of the reactants.
Hint: Answers are a) 16.2 dm^3/mol*h
b) 5.1 × 10^3 s, 2.1 × 10^3 s
Chemistry
1 answer:
andriy [413]2 years ago
3 0

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = -\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B

If C_{A0} and C_{B0} are the inital concentrations and x the concentration reacted at time t, so C_A=C_{A0} -x and C_B=C_{B0} -x and the rate at time t is written as:

-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)

Separating variables and integrating

\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt

The integral in left side is solved by partial fractions, it can be used integral tables

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt

Using logarithm properties (ln x - ln y = ln(x/y))

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a C_{B0} of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, C_{A0}=0.075, C_{B0}=0.05, C_{B}=0.02, it implies that the quantity reacted, x, is 0.03 and C_{A}=0.075-x = 0.045. Then, the value of k would be

kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})

k = 16.21 \frac{dm^3}{mol*1h}

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt

but suppossing that C_A= C_{A0}/2=0.0375 so  C_B=C_{B0}- C_{A0}/2=0.0125, k=16.2 and the same initial concentrations. Replacing in the equation

t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})

t=1.71 h = 1.71*3600 s = 6.1*10^3 s  

For reactant B, C_B= C_{B0}/2=0.025 so  C_A=C_{A0}- C_{B0}/2=0.05, k=16.2 and the same initial concentrations. Replacing in the equation

t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})

t=0.71 h = 0.7*3600 s = 2.5*10^3 s  

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

You might be interested in
Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
coldgirl [10]

The final temperature, t₂ = 30.9 °C

<h3>Further explanation</h3>

Given

24.0 kJ of heat = 24,000 J

Mass of calorimeter = 1.3 kg = 1300 g

Cs = 3.41 J/g°C

t₁= 25.5 °C

Required

The final temperature, t₂

Solution

Q = m.Cs.Δt

Q out (combustion of compound) = Q in (calorimeter)

24,000 = 1300 x 3.41 x (t₂-25.5)

t₂ = 30.9 °C

3 0
3 years ago
Read 2 more answers
A piece of copper is 2x2x4 cm and has a mass of 142, what is the density of copper
ohaa [14]

Answer:

Demisty = mass/volume = 142/ (2×2×4) = 142 / 16 =

8.875g/cm^3

Explanation:

3 0
3 years ago
Why are metal containers not used for storing acids​
kaheart [24]

<em><u>Answer </u></em><em><u>:</u></em><em><u> </u></em><em><u>M</u></em><em><u>etal containers</u></em><em><u> </u></em><em><u>are</u></em><em><u> not</u></em><em><u> </u></em><em><u> </u></em><em><u>used for storing acid</u></em><em><u> because most of the time acid reacts with almost every metal and produces </u></em><em><u>salts</u></em><em><u> or oxid</u></em><em><u>e</u></em><em><u>s</u></em><em><u> </u></em><em><u>which alters the acid characteristics making it useless</u></em><em><u> </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

3 0
2 years ago
PLEASE HELP ME ASAP URGENT IM STRESSED AND STUCK
Genrish500 [490]
Is there any more info lol
But from thinking I got A=DH^2*BC
8 0
2 years ago
Select the correct answer.
almond37 [142]

Answer:

D. 5 moles

Explanation:

C3H8 + 5O2 → 4H2O + 3CO2

            5 mol                 3 mol

So, to make 3 mol CO2    5 mol O2 are needed.

3 0
3 years ago
Other questions:
  • The nucleus is a round, double-walled structure often found in the center of both plant and
    13·2 answers
  • Which particle is j.j thomson credited with discovering?
    15·1 answer
  • CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the s
    13·1 answer
  • What is the independent variable of the sentence "if diet affects weight, then people who eat less will lose weight."?
    6·1 answer
  • K2SO4(aq) + Ba(NO3)2(aq) → 2KNO3(aq) + BaSO4(s)
    5·2 answers
  • The bomb calorimeter in Exercise 102 is filled with 987g water. The initial temperature of the calorimeter contents is 23.32. A
    11·1 answer
  • Please answer asap!
    6·1 answer
  • Identify one trait in a plant that could be changed with selective breeding and explain how the change would be useful? Science.
    5·1 answer
  • Select the correct answer.
    7·2 answers
  • Please help me find the answer , thank you.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!