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BARSIC [14]
1 year ago
13

A colored dye compound decomposes to give a colorless product. The original dye absorbs at 608 nm and has an extinction coeffici

ent of 4.7 ×10⁴ M⁻¹cm⁻¹ at that wavelength.
You perform the decomposition reaction in a 1-cm cuvette in a spectrometer and obtain the following data:
Time (min) - Absorbance at 608 nm
0 - 1.254
30 - 0.941
60 - 0.752
90 - 0.672
120 - 0.545
From these data, determine the rate law for the reaction "dye → product" and determine the rate constant.

Chemistry
1 answer:
dusya [7]1 year ago
6 0

Answer:

First Order reaction

k=  0.0067

Explanation:

To determine the rate law we will have to study the dependence of the rate of the reaction with respect to time.

So we need to calculate the concentrations given the data since  A= ε c l

where ε = molar absorptivity, c is the concentration moles/L, and l is the length of the cuvette.

c = A/εl

Time, min    M, (mol/L) x 10⁻⁵

0                         2.67

30                       2.00

60                       1.60

90                       1.43

120                      1.16

If the reaction were first order, the rate will have to remain constant which is not the case here since the rate of dissaperance is chnging:

Δmol/L/ΔT = (2.00 - 2.67 ) mol /30 min = -0.023 mol/Lmin ( between t= 30 and 0 min)

Δmol/L/ΔT = ( 0.160 - 2) /30 min =  = -0.061 mol/Lmin

If the rate is first order we will have to use the integrated rate law:

ln (A) t  = -kt + ln (A)₀

where (A) is the concentration, t is the time and k is the rate constant

Notice the equation  is of the form y = mx + b where

m= rate constant and b  (A)₀

To verify if our reaction is first order, we will graph the graph the data and see if it is a straght line and and if it is compute the slope which is k.

We could also perform a linear regression with an apprpiat program and obtain the results.

Working in excel the following results were obtained:

y = -0.0067 x  -10.582

k = 0.0067/ min

R² = correlation coefficient = 0.979

First Order reaction

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Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
Grace [21]

Answer : The value of K_{eq} is, 11.2

The value of \Delta G_{rxn} is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

\Delta G^o=-RT\times \ln K_{eq}

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol

Therefore, the value of \Delta G_{rxn} is -9.04 kJ/mol

3 0
2 years ago
Find the molarity of 750 ml solution containing 346 g of potassium nitrate
Zinaida [17]
Given mass of KNO₃=346g
Molar mass of KNO₃=(39.098)+(14)+(15.99*3)=101.068gmol⁻¹
Volume of Solution=750ml=0.75dm³

Molarity=(mass of solute/molar mass of solute)*(1/volume of sol. in dm³)
            =(346/101.068)*(1/0.75)
            =4.56 mol dm⁻³
5 0
2 years ago
How much heat is required to heat 9.61g of ethanol (CH3CH2OH) from 24.10C to 67.30C?
Licemer1 [7]

Answer:

a.)  

To warm the liquid from 35°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - 35) = 4154 J

To vaporize the liquid at 78°C:

(38.56 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 35.154 kJ

Total:

4.154 kJ + 35.154 kJ = 39.3 kJ

b.)  

To warm the solid from -155°C to -114°C:

(0.97 J/g-K) x (42.0 g) x (-114°C - (-155°C)) = 1670 J

To melt the solid at -114°C:

(5.02 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 4.5766 kJ  

To warm the liquid from -114°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - (-114)) = 18547 J

To vaporize the liquid at 78°C:

35.154 kJ  (as in part a.)

Total:

1.670 kJ + 4.5766 kJ + 18.547 kJ + 35.154 kJ = 59.9 kJ

Explanation:

8 0
2 years ago
When a 375 mL sample of nitrogen is kept at constant temperature, it has a pressure of 1.2 atmospheres. What pressure does it ex
kolezko [41]

Answer:

It exerts a pressure of 3.6 atm

Explanation:

This is a gas law problem. We are looking at volume and pressure with temperature being kept constant, thus, the gas law to use is Boyle’s law. It states that at a given constant temperature, the volume of a given mass of gas is inversely proportional to the pressure of the gas.

Mathematically; P1V1 = P2V2

Let’s identify the parameters according to the question.

P1 = 1.2 atm

V1 = 375mL

P2 = ?

v2 = 125mL

We arrange the equation to make room for P2 and this can be written as:

P2 = P1V1/V2

P2 = (1.2 * 375)/125

P2 = 3.6 atm

3 0
1 year ago
Read 2 more answers
A substance occupies one half of an open container. The atoms of the substance are closely packed but are still able to slide pa
SCORPION-xisa [38]

For a substance to occupy one half of an open container with the atoms being able to slide past each other, the most likely phase of the substance would be liquid.

Atoms of gases would not occupy just one-half of an open container because atoms of gases will diffuse and spread out from an open container.

Atoms of solids are fixed about a particular position and will most likely not be able to slide past each other in an open container.

Only the atoms of liquid take the shapes of their containers and do not have the capacity to diffuse out ordinarily. They are also able to flow despite the closeness of the atoms and would occasionally slide past each other.

More on states of matter can be found here: brainly.com/question/9402776

8 0
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