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DiKsa [7]
3 years ago
12

Select all of the following statements that are false about ΔGo and ΔG:a) If the reaction has a large negative ΔGo value, the re

action must reach equilibrium at a small extent of reaction value.b) ΔGo and ΔG can have different values, they don't even have to have the same sign.c) For a reaction that reaches equilibrium, the minimum value of free energy must be at the equilibrium point.d) ΔGo and ΔG have the same magnitude, they just have opposite signs.e) If ΔGo, measured at an extent of reaction = 0.5, is positive, the sign for ΔG when the extent of reaction = 0.80 is also positive.
Chemistry
1 answer:
OLEGan [10]3 years ago
7 0

Answer:

a) If the reaction has a large negative ΔGo value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔGo and ΔG have the same magnitude, they just have opposite signs.

Explanation:

The fraction of the total heat energy if a system that does useful work is known as Gibb's free energy (G) and the change from the initial to final state is designated by \Delta G. It is observed that the values of \Delta G changes with experimental conditions such as temperature , pressure , concentration etc.

\Delta G^0 is the standard free energy change which is a balance of two natural tendencies of any system.

  1. Minimization of potential energy or enthalpic factor \Delta H^0

Maximization of disorderliness or entropic factor T\Delta S^0

Mathematically; \Delta G = \Delta H^0 - T\Delta S^0

Thus; from above mentioned, the statements that are true about ΔG⁰ and ΔG are:

ΔG⁰ and ΔG can have different values, they don't even have to have the same sign

For a reaction that reaches equilibrium, the minimum value of free energy must be at the equilibrium point

If ΔG⁰ , measured at an extent of reaction = 0.5, is positive, the sign for ΔG when the extent of reaction = 0.80 is also positive.

while the false statements include:

a) If the reaction has a large negative ΔG⁰ value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔG⁰ and ΔG have the same magnitude, they just have opposite signs.

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2. what volume of 0.0150-m hcl solution is required to titrate 150. ml of a 0.0100-m ca(oh)2 solution? (10 points)
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100ml volume of  0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.

Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,

    C1 V1 =C2 V2

where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1)  HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1)  Caoh2 solution.

While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give  in the expression we get the volume of the final concentration.

          V1=  C2 V2/ C1

               = 0.0100m . 150ml /0.0150M

               = 100ml

The volume of the hcl solution is 100ml.

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brainly.com/question/7208939

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3 0
1 year ago
An ideal gas is brought through an isothermal compression process. The 3.00 mol of gas goes from an initial volume of 261.6×10−6
Eduardwww [97]

Answer : The temperature and the final pressure of the gas is, 586.83 K and 1.046\times 10^{9}atm respectively.

Explanation : Given,

Initial volume of gas = 261.6\times 10^{-6}m^3

Final volume of the gas = 138.2\times 10^{-6}m^3

Heat released = -9340 J

First we have to calculate the temperature of the gas.

According to the question, this is the case of isothermal reversible compression of gas.

As per first law of thermodynamic,

\Delta U=q+w

where,

\Delta U = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

q=-w

Thus, w = -q = 9340 J

The expression used for work done will be,

w=nRT\ln (\frac{V_2}{V_1})

where,

w = work done = 9340 J

n = number of moles of gas  = 3 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = ?

V_1 = initial volume of gas

V_2 = final volume of gas

Now put all the given values in the above formula, we get the temperature of the gas.

9340J=3mole\times 8.314J/moleK\times T\times \ln (\frac{261.6\times 10^{-6}m^3}{138.2\times 10^{-6}m^3})

T=586.83K

Now we have to calculate the final pressure of the gas by using ideal gas equation.

PV=nRT

where,

P = final pressure of gas = ?

V = final volume of gas = 138.2\times 10^{-6}m^3=138.2\times 10^{-9}L

T = temperature of gas = 586.83 K

n = number of moles of gas = 3 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

P\times (138.2\times 10^{-9}L)=3mole\times (0.0821L.atm/mole.K)\times (586.83K)

P=1.046\times 10^{9}atm

Therefore, the temperature and the final pressure of the gas is, 586.83 K and 1.046\times 10^{9}atm respectively.

6 0
3 years ago
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