Chemistry MCQs for NEET — Practice Questions with Answers

The NCERT states, 'it must be remembered that for the existence of heterogeneous equilibrium pure solids or liquids must also be present (however small the amount may be) at equilibrium, but their concentrations or partial pressures do not appear in the expression of the equilibrium constant' because they are constant. This is also implicitly shown in the $CaCO_3$ decomposition example.

The NCERT explicitly states, 'An equilibrium constant does not give any information about the rate at which the equilibrium is reached.' However, it can predict the extent and direction of a reaction, and calculate equilibrium concentrations, and is temperature dependent.

The NCERT states that the direction of reaction can be predicted by the reaction quotient ($Q_c$) which is equal to $K_c$ at equilibrium. If $Q_c < K_c$, the reaction will proceed in the forward direction to reach equilibrium. Here, $1.0 \times 10^{40} < 2.4 \times 10^{47}$, so the reaction will proceed in the forward direction.

The NCERT states, 'The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.'

A $K_c$ value of 1.0 falls between $10^{-3}$ and $10^3$. As per the NCERT, in this range, 'the concentrations of the reactants and products are comparable, when the system is in equilibrium.'

The NCERT lists as an important feature: 'Expression for equilibrium constant is applicable only when concentrations of the reactants and products have attained constant value at equilibrium state.'

According to the NCERT text, 'When a haloalkane with $\beta$-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from $\beta$-carbon and a halogen atom from the $\alpha$-carbon atom.' Thus, the presence of a $\beta$-hydrogen atom is essential for elimination.

The NCERT text states, 'When a haloalkane with $\beta$-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from $\beta$-carbon and a halogen atom from the $\alpha$-carbon atom.' Alcoholic KOH functions as a strong base, promoting the elimination of a $\beta$-hydrogen and a halogen, leading to an alkene.

As per the NCERT definition, 'Carbon on which halogen atom is directly attached is called $\alpha$-carbon and the carbon atom adjacent to this carbon is called $\beta$-carbon.' In $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$, the bromine is attached to C1, making it the $\alpha$-carbon. The carbon adjacent to C1 (C2) is the $\beta$-carbon.

For an elimination reaction to occur, there must be at least one $\beta$-hydrogen atom. Bromomethane ($\text{CH}_3\text{Br}$) has only an $\alpha$-carbon (the carbon bonded to bromine) but no $\beta$-carbon, and therefore no $\beta$-hydrogen atoms. All other options have $\beta$-hydrogens.