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# in which of the following reactions standard reaction entropy change is positive

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### Mohammed

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## In which of the following reactions, standard reaction entropy change Δ S0 is positive and standard Gibbs's energy change Δ G0 decreases sharply with increasing temperature?

In which of the following reactions, standard reaction entropy change Δ S0 is positive and standard Gibbs's energy change Δ G0 decreases sharply with increasing temperature?

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In which of the following reactions, standard reaction entropy change Δ S0 is positive and standard Gibbs's energy change Δ G0 decreases sharply with increasing temperature?

Question

In which of the following reactions, standard reaction entropy change

( Δ S 0 )

is positive and standard Gibbs's energy change

( Δ G 0 )

decreases sharply with increasing temperature?

A M g ( s ) + 1 2 O 2 ( g ) ⟶ M g O ( s ) B 1 2 C ( graphite ) + 1 2 O 2 ( g ) ⟶ 1 2 C O 2 ( g ) C C ( graphite ) + 1 2 O 2 ( g ) ⟶ C O ( g ) D C O ( g ) + 1 2 O 2 ( g ) ⟶ C O 2 ( g ) Open in App Solution

The correct option is C

C ( graphite ) + 1 2 O 2 ( g ) ⟶ C O ( g ) C ( s ) + 1 2 O 2 ( g ) → C O ( g ) ; Δ η = + 1 2 ;

Also the moles of gases increase and therefore entropy change

( Δ S ) is positive.

An increase in temperature will cause more change in

T Δ S .

Also it is a combustion reaction and thus

Δ H = − v e Since Δ G = Δ H − T Δ S = − v e − ( + v e ) = − v e Suggest Corrections 1

SIMILAR QUESTIONS

Q. In which of the following reactions, standard reaction entropy change

( △ S 0 )

is positive and standard Gibb's energy change

( △ G 0 )

decreases sharply with increasing temperature?

Q. In which of the following reactions, standard reaction entropy charge

( Δ S o )

is positive and standard Gibb's energy charge

( Δ G o )

decreases sharply with increasing temperature?

Q. Standard Gibbs Free energy change

Δ G 0

for a reaction is zero. The value of equilibrium constant of the reaction will be:

Q. For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by:Q. Assertion :If both

Δ H 0 and Δ S 0

are positive then reaction will be spontaneous at high temperature. Reason: All processes with positive entropy change are spontaneous.

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स्रोत : byjus.com

## In which of the following reactions, standard reaction entropy change (S^0) is positive and standard Gibb's energy change (G^0) decreases sharply with increasing temperature?

Click here👆to get an answer to your question ✍️ In which of the following reactions, standard reaction entropy change (S^0) is positive and standard Gibb's energy change (G^0) decreases sharply with increasing temperature?

In which of the following reactions, standard reaction entropy change (△S

Question 0

) is positive and standard Gibb's energy change (△G

0

) decreases sharply with increasing temperature?

AC

(graphite) ​ + 2 1 ​ O 2(g) ​ →CO (g) ​

BCO

(g) ​ + 2 1 ​ O 2(g) ​ →CO 2(g) ​

CMg

(s) ​ + 2 1 ​ O 2(g) ​ →MgO (s) ​

D

2 1 ​ C (graphite) ​ + 2 1 ​ O 2(g) ​ → 2 1 ​ CO 2(g) ​ Hard Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is A)

The change in gaseous moles Δn

g ​

must be highest which indicates the more gaseous product is formed which increases randomness and thus entropy increases.

(A) Δn g ​ =1− 2 1 ​ = 2 1 ​ (B) Δn g ​ =1− 2 3 ​ = 2 −1 ​ (C) Δn g ​ =0− 2 1 ​ = 2 −1 ​ (D) Δn g ​ = 2 1 ​ − 2 1 ​ =0 Hence A is correct.

Solve any question of Chemical Thermodynamics with:-

Patterns of problems

>

89 13

स्रोत : www.toppr.com

## C (graphite) +(1)/(2)O(2)(g) to CO(g)

Among the given reactions only in the case of C("graphite")+(1)/(2) (g) to CO(g) entropy increases because randomness (disorder) increases. Thus, standard entropy change (DeltaS^(@)) is positive . Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic , i.e., DeltaH^(@)=ve We know that, DeltaG^(@)=DeltaH^(@)-TDeltaS^(@) DeltaG^(@)a =-ve-T(+ve) Thus, as the temperature increases, the value of DeltaG^(@) decreases.

Home > English > Class 12 > Chemistry > Chapter > Thermodynamics >

In which of the following reac...

In which of the following reactions,standard reaction entropy change

(Δ S ∘ ) (ΔS∘)

is positive and standard Gibb,s energy change

(Δ G ∘ ) (ΔG∘)

decreases sharply with increasing temperature?

Updated On: 27-06-2022

00 : 30

Text Solution Open Answer in App A C (graphite) + 1 2 O 2 (g)→CO(g) +12O2(g)→CO(g) B CO(g)→ 1 2 O 2 (g)→C O 2 (g)

CO(g)→12O2(g)→CO2(g)

C Mg(s)→ 1 2 O 2 (g)→MgO(s)

Mg(s)→12O2(g)→MgO(s)

D 1 2 C 12C (graphite) + 1 2 O 2 (g)→ 1 2 C O 2 (g) +12O2(g)→12CO2(g) Answer

Solution

Among the given reactions only in the case of

C(graphite)+ 1 2 (g)→CO(g)

C(graphite)+12(g)→CO(g)

entropy increases because randomness (disorder) increases. Thus, standard entropy change (

Δ S ∘ ) ΔS∘)

is positive . Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic , i.e.,

Δ H ∘ =ve ΔH∘=ve We know that, Δ G ∘ =Δ H ∘ −TΔ S ∘ ΔG∘=ΔH∘-TΔS∘ Δ G ∘ a=−ve−T(+ve) ΔG∘a=-ve-T(+ve)

Thus, as the temperature increases, the value of

Δ G ∘ ΔG∘ decreases.

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

hello everyone we have this question which we have to give that in which of the following reactions standard reaction entropy change is and standard Gibbs energy change decreases sharply with increasing temperature for Delta is to be positive that is the reaction in profit with positive there should be increase in the number of gases small letters in which reaction that is happening for we have carbon graphite that is solid plus half O2 which is gaseous form carbon monoxide that is gracious so in the first equation we see that delta in half 21 the Rebel 2 1 - 1 by 2 because this is

the change flight from one from half Glacier small it become one gracious so what will be the change be 1 - 1 by 2 reaction data-ng will get Sri by two gases malls in the reactant side and just one gracious most in the product side so product sagacious minus reactants Idea 1 and 1 by 2 by 2 minus in the third reaction we see that on the product side we have a solid formed service will be zero for solid we have half years more than the reactant side sir

0 - 1 by 2 which is minus one by two similarly we have a check on the 4th one see Delta NG again we have solid so there is no change because half gracious moles are present and half dishes moles are formed so Delta energy is zero so we see that Delta is will be positive only in case of the first one what is the reason for that because the number of gases Moses increasing if the number of guests were small increase the novice Lee the randomness will also increase the directors will become positive now we also have to check whether the standard give energy saving or not with increase in temperature Delta G not is equals to Delta a

- Delta now this is positive we are increasing the temperature so that me this whole thing will become more negative this becomes more negative The Delta G value will become more negative should decrease with increase in temperature isn't it aur we see that Delta G not equal to delta H minus theta dialysis positive temperature is increasing that means this will get an increased value just a minute we will get

increased value if you get this value increases then obviously a larger value will be negative or subtracted from Delta age and when a larger value is subtracted obviously The Delta G will decrease sharply with increase in temperature Soya answer is the first reaction all of it depends upon the number of glacier smoke from the reactant to product thank you

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Mohammed 8 day ago

Guys, does anyone know the answer?