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Here's my explanation
According to Ohm's law the current passed through a material is directly proportional to the voltage.
You can see the mathematical equation that describes this proportionality
The other equation
Where E is electric field at the given location
J is current density
σ (sigma) is a material-dependent parameter called the conductivity
R is the electrical resistance of a uniform specimen of the material
A is the cross-sectional area of the specimen
The resistance of a given material increases with length, but decreases with increasing cross-sectional area
ρ depends on the conductor and is a constant
Here's a table of ρ of the conductor at
Note that ρ is different for conductors at different temperature
ρ increases as the temperature increases
ρ decreases as the temperature decreases
ρ is resistivity
There are many version of the Ohm's law
Now lets come to the main problem.
Where Q is charge
PE is energy
By this you can understand that Voltage is proportional to charge which is proportional to coulombs
There is a relationship between V and C
As the V increases and the Q is thought to be constant the C increases. Means more voltage more coulombs
Now let me show you an example
Net ionic reaction
Each mole of
The whole reaction
As 2C/s is applied for 2s
Use the Faraday's constant to calculate the moles of electrons lost and gained in total.
1 mole of
Thus moles of
Now see what happens if we apply more charge
Consider the charge as 3C/s for 2s
Do the same calculations
As 3C/s is applied for 2s
Thus more the current,voltage and coulomb the more the mol of copper formed thus more the mass of copper formmed
Plotting a graph
If my handwriting is too bad
y = mol of Cu formed from reduction of
x = mol of electrons used in the process
I will stick to diatomic molecules for simplicity. Some general steps are:
Some general rules or tips are:
Let's construct an MO diagram for
CHOOSE YOUR ATOMIC ORBITALS & AXES
ATOMIC ORBITAL ENERGY DIAGRAM
From the original atomic orbital energies, we then construct the two atomic orbital energy diagrams:
(On an exam, you may only be expected to work with homonuclear diatomic molecules, in which case you probably don't have to worry about relative energies.)
GENERATION OF MOLECULAR ORBITALS
Then, each pair of atomic orbitals generates molecular orbitals as follows. Here is an
For lithium through nitrogen (including nitrogen), orbital mixing occurs, so that the
Therefore, carbon has its energy ordering switched compared to oxygen, and so, the
OVERALL MO DIAGRAM
Now put it together to get:
Note that on an exam, you probably aren't expected or required to know exactly how the interactions go. It's enough to know the relative energies of the molecular orbitals compared to the atomic orbitals, and to know how many valence electrons go in.
DISCLAIMER: LONG ANSWER!
#"NAD"^(+)(aq) + "H"^(+)(aq) + 2e^(-) -> "NADH"(aq)#
The general Nernst equation relates the non-standard
#E_"cell" = E_"cell"^@ - (RT)/(nF)lnQ#
We now know how to write
#Q = (["NADH"])/(["NAD"^(+)]["H"^(+)])#
Hence, the Nernst equation for this problem, which is for a half-cell, is:
#bb(E_"red" = E_"red"^@ - (RT)/(nF)ln((["NADH"])/(["NAD"^(+)]["H"^(+)]))#
We are also given
So, we just use the equation we wrote down in part
#color(blue)(E_"red") = -"0.11 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M")/("1 M"cdot 10^(-7) "M"))#
#= color(blue)(-"0.32 V")#
This indicates that the reduction of
Since the important reaction is the reduction of the iron center, let's focus on that half-reaction. However, don't forget our other half-reaction. Since
Thus, we flip our first half reaction, which was a reduction to begin with, and turn it into an oxidation half-reaction.
#2("Fe"^(3+)(aq) + cancel(e^(-)) -> "Fe"^(2+)(aq))#
#"NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + cancel(2e^(-))#
#color(blue)(2"Fe"^(3+)(aq) + "NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + 2"Fe"^(2+)(aq))#
I'll leave you to check that the mass and charge are balanced.
#"Fe"^(3+)(aq) + e^(-) -> "Fe"^(2+)(aq)#, #E_"red"^@ = "0.25 V"#
We now essentially repeat what we did in parts
This is a pretty bulky reaction quotient expression...
#Q = (["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"])#
Thus, the Nernst equation for this problem is:
#E_"cell" = E_"cell"^@ - (RT)/(nF)ln((["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"]))#
We first have to figure out
Since we know from part
#E_"cell"^@ = E_"red"^@ + E_"ox"^@#
#= "0.25 V" + [-stackrel(E_"red"^@)overbrace((-"0.11 V"))]#
#=# #+"0.36 V"#
Now, we can calculate the non-standard cell potential at
#color(blue)(E_"cell") = "0.36 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M"cdot10^(-7)"M"cdot("0.004 M")^2)/(("0.01 M")^2cdot"1 M"))#
#= color(blue)(+"0.59 V")#
Thus, the reduction of the iron center in cytochrome c by
The potential E for a 1/2 cell is given by:
At 298K this can be simplified to:
Where z is the no. of moles of electrons transferred which, in this case =2.
Putting in the numbers:
It helps to consider the
The most +ve 1/2 cell is the one that will take in the electrons so you can see that the 1st 1/2 cell will move right to left and the 2nd 1/2 cell will move left to right in accordance with the arrows.
The two 1/2 equations are therefore:
To get the electrons to balance we X the 2nd 1/2 cell by 2 then add:
Now we have to use the full Nernst Equation:
This can be simplified at 298K to:
Where z = 2.
The clearest way is to subtract the least +ve
Using the values in (c):
From the complete equation the reaction quotient Q is written:
pH = 7
Putting the numbers into The Nernst Equation:
The basic formula is
To answer this, we can take a look at the ionic charges of the elements involved.
Since oxygen is in group 16, it will have
Recall that an atom will want to gain or lose electrons to fulfill the octet rule; it wants to do whatever is easiest to obtain
For an oxygen atom, the easiest way to get
For the alkali metals, which are located in group 1 of the periodic table, they each have
In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol
The compound the alkali metal
The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to put the charge of the anion (the negative ion, in this case is
The main postulates or assumptions are (for ideal gases):
As for "one kinetic gas equation", there is not just one. But I can derive the Maxwell-Boltzmann distribution, and from there one could derive many other equations... some of them being:
The following derivation is adapted from Statistical Mechanics by Norman Davidson (1969). Fairly old, but I kinda like it.
Consider the classical Hamiltonian for the free particle in three dimensions:
#H = (p_x^2)/(2m) + (p_y^2)/(2m) + (p_z^2)/(2m)#,
#p = mv#is the momentum and #m#is the mass of the particle.
The distribution function will be denoted as
From Statistical Mechanics by Norman Davidson (pg. 150), we have that
#(dN)/N = (e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n)/zeta#,
#beta = 1//k_BT#is a constant, with #k_B#as the Boltzmann constant and #T#as the temperature in #"K"#.
#H(p_1, . . . , p_n; q_1, . . . , q_n)#is the Hamiltonian for a system with #3N#momentum coordinates and #3N#position coordinates, containing #N#particles in 3 dimensions.
#zeta = int cdots int e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n#is the classical phase integral.
This seems like a lot, but fortunately we can look at three coordinates for simplicity.
A phase integral is an integral over
#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(int_(-oo)^(oo) int_(-oo)^(oo) int_(-oo)^(oo) "exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)#
Now, let's take the bottom integral and separate it out:
#zeta = int_(-oo)^(oo) e^(-p_x^2//2mk_BT)dp_x cdot int_(-oo)^(oo) e^(-p_y^2//2mk_BT)dp_y cdot int_(-oo)^(oo) e^(-p_z^2//2mk_BT)dp_z#
In phase space, each of these integrals are identical, except for the particular notation. These all are of this form, which is tabulated:
#2int_(0)^(oo) e^(-alphax^2)dx = cancel(2 cdot 1/2) (pi/alpha)^(1//2)#
Notice how the answer has no
#zeta = (2pimk_BT)^(3//2)#
So far, we then have:
#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(2pimk_BT)^(3//2)#
Next, it is convenient to transform into velocity coordinates, so let
#(dN(v_x,v_y,v_z; x,y,z))/(N) = (e^(-m^2(v_x^2 + v_y^2 + v_z^2)//2mk_BT)m^3dv_xdv_ydv_z)/(2pimk_BT)^(3//2)#
#= (m/(2pik_BT))^(3//2)e^(-m(v_x^2 + v_x^2 + v_z^2)//2k_BT)dv_xdv_ydv_z#
Now, we assume that the gases are isotropic, meaning that no direction is any different from any other.
To enforce that, suppose we enter velocity space, a coordinate system analogous to spherical coordinates, but with components of
#v_x = vsintheta_vcosphi_v#
#v_y = vsintheta_vsinphi_v#
#v_z = vcostheta_v#
With the differential volume element being
#(dN(v, theta_v, phi_v))/(N)#
#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v cos^2phi_v + v^2 sin^2theta_v sin^2phi_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#
#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v(cos^2phi_v + sin^2phi_v) + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#
#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#
#=> ul((dN(v, theta_v, phi_v))/(N) = (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv sintheta_v d theta_v d phi_v)#
We're almost there! Now, the integral over allspace for
#int_(0)^(pi) sintheta_vd theta_v = 2#
And the integral over allspace for
#int_(0)^(2pi) dphi_v = 2pi#
So, we integrate both sides to get:
#color(blue)(barul(|stackrel(" ")(" "(dN(v))/(N) -= f(v)dv = 4pi (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv" ")|))#
And this is the Maxwell-Boltzmann distribution, the one you see here:
DERIVATION OF SOME OTHER EQUATIONS
This is only a summary.
Gas Speed Equations
#<< v >> = int_(0)^(oo) vf(v)dv = sqrt((8k_BT)/(pim)#
#v_(RMS) = << v^2 >>^(1//2) = (int_(0)^(oo) v^2f(v)dv)^(1//2) = sqrt((3k_BT)/m)#
#v_(mp)#: Take the derivative of #f(v)#, set the result equal to #0#, and solve for #v#to get
#v_(mp) = sqrt((2k_BT)/m)#
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