Answer:
<h2>belajar jangan malas malas</h2>
Answer:
the would test random expirements
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
To calculate the <span>δ h, we must balance first the reaction:
NO + 0.5O2 -----> NO2
Then we write all the reactions,
2O3 -----> 3O2 </span><span>δ h = -426 kj eq. (1)
O2 -----> 2O </span><span>δ h = 490 kj eq. (2)
NO + O3 -----> NO2 + O2 </span><span>δ h = -200 kj eq. (3)
We divide eq. (1) by 2, we get
</span>O3 -----> 1.5O2 δ h = -213 kj eq. (4)
Then, we subtract eq. (3) by eq. (4)
NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)
eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)
O -----> 0.5O2 <span>δ h = -245 kj eq. (6)
</span>
Add eq. (6) to eq. (5), we get
NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj
<em>ANSWER:</em> <em>NO + O ----> NO2 δ h = -232 kj</em>
Answer:
See explanation
Explanation:
If we look at the electron configuration closely, we will discover that the element must have had a ground state electron configuration of 2,4.
This is because, the innermost shell usually holds two electrons while the outer shells hold eight electrons each. The four electrons must be accommodated in the second shell in the ground state configuration of the compound.
However, when the atom is excited, one electron from this shell may move to the third shell to give the excited state configuration 2-3-1 as shown in the question.
