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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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A small amount wet of hydrogen gas can be prepared by the reaction of zinc with excess hydrochloric acid and trapping the gas pr
dybincka [34]

Answer : The partial pressure of the hydrogen is, 705.9 mmHg

Explanation :

According to the Dalton's law of partial pressure,

P_T=P_{H_2}+P_{H_2O}

where,

P_T = total pressure of the gas = 729.7 mmHg

P_{H_2} = partial pressure of the hydrogen gas = ?

P_{H_2O} = partial pressure of the water = 23.8 mmHg (standard value)

Now put all the given values in the above expression, we get:

729.7mmHg=P_{H_2}+23.8mmHg

P_{H_2}=705.9mmHg

Therefore, the partial pressure of the hydrogen is, 705.9 mmHg

4 0
3 years ago
Magnesium sulfate can be made by reacting magnesium metal with an acid . A gas is also produced . Name this gas
jekas [21]

Answer:

the answer to your question is

Explanation:

hydorgen

Mg+ H2SO4 --------> MgSO4 + H2

8 0
3 years ago
What mass of iron is formed when 10 grams of carbon react with 80 grams of iron iii oxide?
yKpoI14uk [10]

Answer:

55.85 grams of Fe is formed.

Explanation:

Identify the reaction:

2Fe₂O₃  +  3C  →  4Fe  +  3CO₂

Identify the limiting reactant, previously determine the mol of each reactant

(mass / molar mass)

10 g / 12 g/m = 0.83 moles C

80 g / 159.7 g /m = 0.500 moles Fe₂O₃

2 moles of oxide need 3 moles of C, to react

0.5 moles of oxide, will need ( 0.5  . 3)/ 2 = 0.751 mol

I have 0.83 moles of C, so C is the excess.

The limiting is the oxide.

3 mol of C need 2 mol of oxide to react

0.83 mol of C, will need (0.83  . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)

Ratio is 2:4 (double)

If I have 0.5 moles of oxide, I will produce the double, in the reaction.

1 mol of Fe, will be produce so its mass is 55.85 g

5 0
3 years ago
The diagram below shows a cell placed in a solution. What will happen to the cell?
drek231 [11]

the cell will expand as water moves into it.

7 0
3 years ago
Help me out please!!
malfutka [58]

Answer:6.48*10^{-2}

Explanation:

alright, dawg, lets get this bread. CHEMISTRY? OH YEAH I LOVE CHEMISTRY.

what is a mol? do you know who avogadro is? sounds like avocado. free shavocado. ok so you MUST REMEMBER THIS NUMBER PLEASE.

please remember this number and commit it to your memory: avogadros number

6.02 * 10^{23}

this is how much a mole is. you know how a pair is 2 and a dozen is 12? ok so a mole is 6.02 * 10^{23} it is confusing at first but hopefully this helps you to understand.

now that we understand this..... lets perform this calculation with a calculator

(3.90 * 10^{22}) / (6.02 * 10^{23})

notice i divide the question by the avogadros number to find out how many moles are in the number. ok but listen... it gets into a tough area here... because HOW ARE WE TO DIVIDE SUCH A HUMONGOUS NUMBER BY ANOTHER HUMONGOUS NUMBER?!?!?

its easy, its cake, just listen this is how you do it. only focus on the numbers NOT the 10 exponential ones. so just 3.90 and 6.02 ok? lets divide these two numbers 3.90 / 6.02 and we get 0.6478... how interesting... ok now lets deal with the exponents of 10. notice that we are DIVIDING these numbers so think of it as subtracting the exponents of ten.....    22 minus 23 equals -1

so we have 0.6478 * 10^{-1}

now this negative 1 thing is annoying so lets just make it to the power of 0

0.06478 * 10^{0}

and anything to the power of 0 just becomes 1.

0.06478

so this is our answer but keep in mind we need 3 sig figs. if we round then we get 0.0648

put this into scientific notation we get 6.48*10^{-2}

5 0
3 years ago
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