What will be the major product of the following sequence of reactions?
\(n−Bu−≡ \xrightarrow[{(ii)Lindlar catalyst, H_2}]{(i)n-BuLi,n-C_5H_{11}Cl}\)
\(n−Bu−≡ \xrightarrow[{(ii)Lindlar catalyst, H_2}]{(i)n-BuLi,n-C_5H_{11}Cl}\)
The Correct Option is C
Solution and Explanation
Hence, the correct option is (C)
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
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Top JEE Main Chemical Reactions Questions
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Assertion A: H2Te is more acidic than H2S.
Reason R: Bond dissociation enthalpy of H2Te is lower than H2S.
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\(\text{(A) } \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}\)
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\(\text{(C) } 2\text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2\)
\(\text{(D) } 2\text{MnO}_4^- + 3\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow 5\text{MnO}_2 + 4\text{H}^+\)
Choose the correct answer from the options given below. - Products for the below reaction are:
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Top JEE Main Questions
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The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
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Concepts Used:
Law of Chemical Equilibrium
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.