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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

For a reaction A → B, E _ a = 10kJ/mol,Δ H = 5kJ/mol. Thus...

For a reaction A→B,Ea=10kJ/mol,\DeltaH=5kJ/mol.\text{A} \to \text{B,}{E}_{a} = 10\text{kJ/mol,}\text{\Delta H} = 5\text{kJ/mol}\text{.} Thus potential energy profile for this reaction isFor a reaction A → B, E _ a = 10kJ/mol,Δ H = 5kJ/mol. Thus potential energy profile for this reaction isFor a reaction A → B, E _ a = 10kJ/mol,Δ H = 5kJ/mol. Thus potential energy profile for this reaction isFor a reaction A → B, E _ a = 10kJ/mol,Δ H = 5kJ/mol. Thus potential energy profile for this reaction isFor a reaction A → B, E _ a = 10kJ/mol,Δ H = 5kJ/mol. Thus potential energy profile for this reaction is

Options

  1. ALet x _ 1 , x _ 2 , , x _ 100 be in an arithmetic progression, with x _ 1 = 2 and their mean equal to 200 . If y _ i =
  2. BLet x _ 1 , x _ 2 , , x _ 100 be in an arithmetic progression, with x _ 1 = 2 and their mean equal to 200 . If y _ i =
  3. CLet x _ 1 , x _ 2 , , x _ 100 be in an arithmetic progression, with x _ 1 = 2 and their mean equal to 200 . If y _ i =
  4. DLet x _ 1 , x _ 2 , , x _ 100 be in an arithmetic progression, with x _ 1 = 2 and their mean equal to 200 . If y _ i =

Correct answer

B. Let x _ 1 , x _ 2 , , x _ 100 be in an arithmetic progression, with x _ 1 = 2 and their mean equal to 200 . If y _ i =

Step-by-step solution

ΔH=(Ea)ƒ–(Ea)b\Delta H = ( {E}_{a}{)}_{ƒ} – ( {E}_{a}{)}_{b}
5=10–(Ea)b5 = 10 – ( {E}_{a}{)}_{b}
(Ea)b=5kJ/mol.{{\text{(E}}_{a} \text{)}}_{b} = 5\text{kJ/mol}\text{.}
(Ea)ƒ>(Ea)b( {E}_{a}{)}_{ƒ} > ( {E}_{a}{)}_{b} hence option (B)\left(B\right) is correct
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