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sladkih [1.3K]
2 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]2 years ago
8 0

Answer:

a. Not change the free energy value

Explanation:

AURORKA [14]2 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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Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra
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Answer:

Mole fraction of methanol will be closest to 4.

Explanation:

Given, Mass of methanol = 128 g

Molar mass of methanol = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{128\ g}{32.04\ g/mol}

Moles\ of\ methanol = 3.995\ mol

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Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

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So, according to definition of mole fraction:

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<u>Mole fraction of methanol will be closest to 4.</u>

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The answer is toluene I think
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By remaining in a stretched condition, the rubber is in a state of high potential energy, when the force holding the rubber in place is removed, due to the laws of thermodynamics, the polymers in the rubber curls back to their state of "random" tangled mass releasing the stored potential energy in the process and doing work such as moving items placed in the rubber's path of motion such as an object that has weight, w then takes up the kinetic energy 1/2×m×v² which can can result in the flight of the object.

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Answer:

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Explanation:

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Also, we need to understand that when the  Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

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e. 6 protons and 5 neutrons  =  Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons  =  Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons  =  Neutron- proton ratio = N/P =  7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons =  Neutron- proton ratio = N/P = 0/1 =0 = unstable

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