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Paul [167]
3 years ago
14

The ΔG of a reaction would be at the minimum when... Group of answer choices the equilibrium constant is equal to 1 (i.e., the r

eactant and product concentrations are always equal). the reactant and product concentrations don't change over time (the system is at equilibrium). the entropy has reached its maximum positive value. the reaction goes to completion. the reaction is very slow
Physics
1 answer:
Blizzard [7]3 years ago
5 0

Answer:

the equilibrium constant is equal to 1 (i.e., the reactant and product concentrations are always equal).

Explanation:

ΔG is a symbol related to Gibbs free energy, which is a physical quantity related to thermodynamics. ΔG refers to the difference between the change in enthalpy (and sometimes entropy) and the temperature of a chemical reaction.

Gibbs free energy is very useful for measuring the work done between the reactants in a reaction. It is calculated using the formula: ΔG = change in enthalpy - (temperature x change in entropy).

The ΔG of a reaction would have a minimum value (zero), if the equilibrium constant is equal to 1 (that is, the concentrations of the reagent and the product are always equal).

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Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

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Where:

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U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

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y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

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f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

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The magnitude of the frictional force between the car and the track is 367.763 N.

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