Using the thermodynamic information in the aleks data tab, calculate the standard reaction free energy of the following chemical
1 answer:
The standard reaction free energy is the total energy change of a chemical reaction that occurs at standard conditions. The standard reaction free energy is calculated by subtracting the heat of formation from the enthalpy of formation.
standard reaction free energy = -Hf-Hf
The standard Gibbs free energy, on the other hand, is the total of the enthalpy and the entropy. The Gibbs free energy is a thermodynamic potential that depends on the enthalpy and entropy of a system. The change in the Gibbs free energy is equal to ΔG=ΔH-TΔS.
The enthalpy of formation is a measure of how much heat must be put in to break down a compound into its constituent elements. The heat of formation is an estimate of how much heat will be released when the same compound is formed from its constituent elements.
<h3>How is the standard reaction free energy calculated?</h3>The standard reaction free energy is the change in Gibbs free energy, which is the enthalpy minus the entropy. This can be calculated by using E=ΔH-TΔS. The ΔG for a reaction with a standard state change from
atm to
K (the most common atmospheric pressure and temperature), and 1 mol of reactants is equal to 0. The ΔG
for a reaction with standard state changes from 1 atm to 298 K and from
to
atm are equal to
.
The standard reaction free energy can be calculated by using the following equation:
E=ΔH-TΔS, where E is the standard free energy of the reaction.
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4. The Coyote has an initial position vector of .
4a. The Coyote has an initial velocity vector of . His position at time
is given by the vector
where is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with
The Coyote hits the ground when :
4b. Here we evaluate at the time found in (4a).
5. The shell has initial position vector , and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:
Answer:
a) Ffr = -0.18 N
b) a= -1.64 m/s2
c) t = 9.2 s
d) x = 68.7 m.
e) W= -12.4 J
f) Pavg = -1.35 W
g) Pinst = -0.72 W
Explanation:
a)
- While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
- This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
- So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.
where μk is the kinetic friction coefficient, and Fn is the normal force.
- Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:
- We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:
b)
- According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
- In this case, this net force is the friction force which we have already found in a).
- Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
- We can write the expression for a as follows:
c)
- Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:
- Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:
d)
- From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
- So, we can use the following kinematic equation in order to find the displacement before coming to rest:
- Since the puck comes to a stop, vf =0.
- Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:
e)
- The total work done by the friction force on the object , can be obtained in several ways.
- One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
- Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:
f)
- By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
- If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:
g)
- The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
- Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows: