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sladkih [1.3K]
3 years ago
5

For the reaction A + B − ⇀ ↽ − C + D A+B↽−−⇀C+D , assume that the standard change in free energy has a positive value. Changing

the conditions of the reaction can alter the value of the change in free energy ( Δ G ) (ΔG) . Classify the conditions as to whether each would decrease the value of Δ G ΔG , increase the value of Δ G ΔG , or not change the value of Δ G ΔG for the reaction. For each change, assume that the other variables are kept constant.a. Adding a catalystdecrease the free energy value, increase the free energy value, or not change the free energyb. increasing [C] and [D]decrease the free energy value, increase the free energy value, or not change the free energyc. Coupling with ATP hydrolysisdecrease the free energy value, increase the free energy value, or not change the free energyd.Increasing [A] and [B]decrease the free energy value, increase the free energy value, or not change the free energy
Chemistry
2 answers:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

a. Not change the free energy value

Explanation:

AURORKA [14]3 years ago
4 0

Answer:

a. Not change the free energy value

b. Increase the free energy value

c. Decrease the free energy value

d. Decrease the free energy value

Explanation:

a. Adding a catalyst:

A catalyst is a substance that will reduce the activation energy of a reaction, it means that the reaction will occur fast. The values of enthalpy, entropy, and free energy are not affected by a catalyst, so ΔG remains the same.

b. Increasing [C] and [D]:

For a reversible reaction, the value of free energy can be calculated by:

ΔG = ΔG° + RT*lnK

Where ΔG° is the standard value for free energy, R is the gas constant, T is the temperature, and K is the constant of equilibrium, which in this case:

K = ([C]*[D])/([A]*[B])

When [C] and [D] increase, the value of K increases, and lnK also increases, then, the value of ΔG increases.

c. Coupling with ATP hydrolysis:

The free energy can be calculated by:

ΔG = ΔH - TΔS

Where ΔH is the change in enthalpy, and ΔS the change in entropy. The ATP hydrolysis is an exothermic reaction, so ΔH <0. When it is coupled, it will reduce the total value of ΔH, and because of that, the value of ΔG will decrease.

d. Increasing [A] and [B]:

As explained above, the increasing at [A] and [B] will decrease the value of K, so the value of lnK will decrease, and ΔG value will also decrease.

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6 0
3 years ago
Read 2 more answers
1/ Which graph illustrates constant speed and velocity
Maksim231197 [3]
Velocity is said to be constant if its magnitude as well direction at any instant is remains the same. In D, if you draw a line parallel to y-axis at any time t, you can see that velocity is same. Hence, D is the correct graph.

The kinetic energy of gaseous molecules is greater than that of liquid molecules. Therefore, in gas, kinetic energy overcomes the force of attraction between molecules. In short, in gas phase, particles move at high speed and hence they have less force of attraction. In case of liquid phase, particles are close enough as a result there is much more force of attraction compared to gaseous molecules. In liquid state, kinetic energy cannot overcome force of attraction therefore, liquid molecules slow down.

Therefore, B is the correct answer.
4 0
3 years ago
Calculate the pOH of a solution if the concentration of hydroxide ions (OH-) is 1.9 x 10-5M?
OLEGan [10]

Answer:

9.28

Explanation:

pOH refers to a measure of hydroxide ions concentration. pOH tells about the alkalinity of a solution. If pOH is less than 7 then aqueous solutions are alkaline, acidic if pOH is greater than 7 and neutral if pOH is equal to 7.

Concentration of the hydroxide ions = 1.9 x 10-5 M

pH = -log(1.9\times 10^{-5})=4.72

pOH = 14 - pH

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6 0
3 years ago
Consider a 1260-kg automobile clocked by law-enforcement radar at a speed of 85.5 km/h. If the position of the car is known to w
-BARSIC- [3]
The expected speed is v = 85.5 km/h
 v = 85.5 km/h = (85.5 km/h)*(0.2778 (m/s)/(km/h)) = 23.75 m/s

If there is an uncertainty of 2 meters in measuring the position, then within a 1-second time interval:
The lower measurement for the speed is v₁ = 21.75 m/s,
The upper measurement for the speed is v₂ = 25.75 m/s.
The range of variation is
Δv = v₂ - v₁ = 4 m/s

The uncertainty in measuring the speed is
Δv/v = 4/23.75 = 0.1684 = 16.84%

Answer: 16.8%
3 0
3 years ago
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