JEE Exam > JEE Questions > Consider the following reaction:CaCO3(s)CaO(s...
Start Learning for Free
Consider the following reaction:
CaCO3(s) ⇌ CaO(s) + CO2(g)
The decomposition of limestone is non-spontaneous at 298 K.
The ΔHº and ΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.
At what temperature will the decomposition become spontaneous?
CaCO3(s) ⇌ CaO(s) + CO2(g)
The decomposition of limestone is non-spontaneous at 298 K.
The ΔHº and ΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.
At what temperature will the decomposition become spontaneous?
- a)1000 K
- b)Below 500oC
- c)500oC
- d)Above 827oC
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decompositi...
According to Gibb's equation:
The reaction becomes spontaneous at a temperature at which
is a negative value.
At 298K:
Since ΔGº is positive, the reaction is not spontaneous at this temperature.
Hence, the reaction is spontaneous above 1100 K or 827oC.
The reaction becomes spontaneous at a temperature at which
is a negative value.At 298K:
Since ΔGº is positive, the reaction is not spontaneous at this temperature.
Hence, the reaction is spontaneous above 1100 K or 827oC.
Free Test
FREE
| Start Free Test |
Community Answer
Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decompositi...
Decomposition of limestone is non-spontaneous at 298 K because it requires an input of energy to break the bonds in the CaCO3 molecule and produce CaO and CO2. This energy input is typically provided through heating, which increases the temperature of the system and provides the necessary activation energy for the reaction to occur.
At lower temperatures, such as 298 K, the average kinetic energy of the molecules is not sufficient to overcome the activation energy barrier, and the reaction does not proceed spontaneously. This means that if you were to simply place a sample of limestone at 298 K, it would remain as CaCO3 and not decompose into CaO and CO2.
However, if the temperature is increased to a value above the activation energy, the reaction becomes spontaneous and the decomposition of limestone occurs. The increased kinetic energy of the molecules allows them to overcome the energy barrier and break the bonds in the CaCO3 molecule, leading to the formation of CaO and CO2.
In summary, the non-spontaneity of the decomposition of limestone at 298 K is due to the lack of sufficient energy to overcome the activation energy barrier.
At lower temperatures, such as 298 K, the average kinetic energy of the molecules is not sufficient to overcome the activation energy barrier, and the reaction does not proceed spontaneously. This means that if you were to simply place a sample of limestone at 298 K, it would remain as CaCO3 and not decompose into CaO and CO2.
However, if the temperature is increased to a value above the activation energy, the reaction becomes spontaneous and the decomposition of limestone occurs. The increased kinetic energy of the molecules allows them to overcome the energy barrier and break the bonds in the CaCO3 molecule, leading to the formation of CaO and CO2.
In summary, the non-spontaneity of the decomposition of limestone at 298 K is due to the lack of sufficient energy to overcome the activation energy barrier.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer?
Question Description
Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer?.
Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer?.
Solutions for Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider the following reaction:CaCO3(s)CaO(s) + CO2(g)The decomposition of limestoneis non-spontaneous at 298 K.TheΔHºandΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.At what temperature will the decomposition become spontaneous?a)1000 Kb)Below 500oCc)500oCd)Above 827oCCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup