Chemistry, 26.11.2019 07:31 Tips101

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 0.050 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

Answers: 2

Show answers

Answers

Answer from: ameliaxbowen7

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)

Answer from: Quest

your mom.com . : 0000

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

Another question on Chemistry

Chemistry, 22.06.2019 02:40

How many liters of hydrogen gas will be produced at stp from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phosphoric acid (h3po4) the equation is 3mg + 2h3(> mg(po4)2+3h2

Answers: 1

Answer

Chemistry, 22.06.2019 05:50

What are transitions between a liquid and a solid called? identify which way they are transitioning

Answers: 2

Answer

Chemistry, 22.06.2019 07:30

Label a-f based on the table using c for concentrated and d for dilute

Answers: 2

Answer

Chemistry, 22.06.2019 11:00

Problem page combustion of hydrocarbons such as pentane ( c5 h12 ) produces carbon dioxide, a "greenhouse gas." greenhouse gases in the earth's atmosphere can trap the sun's heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.(a) write a balanced chemical equation, including physical state symbols, for the combustion of liquid pentane into gaseous carbon dioxide and gaseous water. (b) suppose 0.350 kg of pentane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 degree c. calculate the volume of carbon dioxide gas that is produced.be sure your answer has the correct number of significant digits.

Answers: 2

Answer

You know the right answer?

a second- order reaction of the type a + b --> p was carried out in a solution that was initially...

Questions