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astraxan [27]
3 years ago
13

Which of the following statements is true? A) This reaction will be spontaneous only at high temperatures. B) This reaction will

be spontaneous at all temperatures. C) This reaction will be nonspontaneous at all temperatures. D) This reaction will be nonspontaneous only at high temperatures. E) It is not possible to determine without more information, 8. (2) Identify the compound with the standard free energy of formation equal to zero. A) KCI(S) B) HF(g) C) NO(g) D) Al(s) E) It is hard to determine.
Chemistry
1 answer:
viktelen [127]3 years ago
7 0

Answer:

D) This reaction will be nonspontaneous only at high temperatures.

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

  • ΔG < 0 , the reaction is Spontaneous
  • ΔG > 0 , the reaction is non Spontaneous
  • ΔG = 0 , the reaction is at equilibrium

The reaction has the value for ∆H = negative , and ∆S = negative ,

Now ,

∆G = ∆H -T∆S

     = ( - ∆H ) - T( - ∆S )

     =  ( - ∆H ) +T(  ∆S )

Now, for making the reaction Spontaneous ΔG = negative ,

Hence ,

The temperature is low, then the value for  ΔG will be negative , i.e. , Spontaneous reaction .

And , vice versa , at higher temperature , the reaction will have ΔG positive , and the reaction will be non -Spontaneous reaction .

The standard free energy of formation will be zero , only for the compounds that are in their pure form ,

Hence , Al(s) will have ΔG = 0 .

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
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Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

7 0
3 years ago
Determine the length of the object shown.<br><br><br><br> 5.6 cm<br> 5.60 cm<br> 5.600 cm<br> 6 cm
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Salutations!

Determine the length of the object shown below.

The length of the scale is 5.6 cm.

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