NEET Practice Questions (MCQs) with Answers & Solutions

According to the NCERT text, 'If $K_c > 10^3$, products predominate over reactants, i.e., if $K_c$ is very large, the reaction proceeds nearly to completion.' A value of $2.4 \times 10^{47}$ is significantly greater than $10^3$.

The NCERT text states: 'The numerical value of the equilibrium constant for a reaction indicates the extent of the reaction. But it is important to note that an equilibrium constant does not give any information about the rate at which the equilibrium is reached.' Therefore, a large $K_c$ indicates extensive product formation at equilibrium, not a slow rate.

As per the NCERT text, 'If $K_c < 10^{-3}$, reactants predominate over products, i.e., if $K_c$ is very small, the reaction proceeds rarely.' A value of $5.0 \times 10^{-5}$ is smaller than $10^{-3}$.

The NCERT states, 'If $K_c > 10^3$, products predominate over reactants, i.e., if $K_c$ is very large, the reaction proceeds nearly to completion.' A $K_c$ value of $5.4 \times 10^{18}$ is very large, indicating a strong preference for product formation at equilibrium.

The NCERT implies that for intermediate $K_c$ values (between $10^{-3}$ and $10^3$), the concentrations of reactants and products are comparable. Since $K_c = 0.5$ falls within this range, both reactants and products will be present in comparable amounts.

The NCERT states, 'The equilibrium constant for an exothermic reaction (negative $\Delta H$) decreases as the temperature increases.' For the formation of ammonia, $\Delta H$ is negative (exothermic).

The NCERT states, 'Kp = $pCO_2$ = $2 \times 10^5$ Pa/10^5 Pa = 2.00'. This means that $pCO_2$ divided by $10^5$ Pa equals 2.00, so $pCO_2 = 2.00 \times 10^5$ Pa. The example explicitly demonstrates this calculation for $K_p$ in the context.

The NCERT specifies, 'The equilibrium constant for an endothermic reaction (positive $\Delta H$) increases as the temperature increases.'

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.