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Fudgin [204]
2 years ago
12

Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ

ine whether the reaction is first order or second order. determine whether the reaction is first order or second order. this is the first order reaction. this is the second order reaction. submitprevious answers correct part b what is the rate constant?
Chemistry
1 answer:
Whitepunk [10]2 years ago
7 0
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

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93.2 mL of a 2.03 M potassium fluoride (KF) solution
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1.98 M

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Given data

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3 0
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Read 2 more answers
Write a balanced equation for the transmutation that occurs when a scandium-48 nucleus undergoes beta decay.
Tju [1.3M]

Answer:

A scandium-48 nucleus undergoes beta-minus decay to produce a titanium-48 nucleus.

\rm ^{48}_{21}Sc \to ^{48}_{22}Ti + ^{\phantom{1}\,0}_{-1}e^{-} + \bar{\mathnormal{v}}_e.

Explanation:

There are two types of beta decay modes: beta-minus and beta-plus.

In both decay modes, the mass number of the nucleus stays the same.

However, in a beta-minus decay, the atomic number of the nucleus increases by one. In a beta-plus decay, the atomic number decreases by one.

Each beta-minus decay releases one electron and one electron antineutrino. Each beta-plus decay releases one positron and one electron neutrino.

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This question did not specify whether the decay here is beta-plus or a beta-minus. However, the relative atomic mass of this element can give a rough estimate of the mode of decay.

Each element (e.g, \rm Sc) can have multiple isotopes. These isotopes differ in mass. The relative atomic mass of an element is an average  across all isotopes of this element. This mass is weighted based on the relative abundance of the isotopes. Its value should be closest to the most stable (and hence the most abundant) isotope.

The mass number of scandium-48 is significantly larger than the relative atomic mass of this element. In other words, this isotope contains more neutrons than isotopes that are more stable. There's a tendency for that neutron to convert to a proton- by beta-minus decay, for example.

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