JEE Chemistry - MCQ Practice Questions

At the critical point, the compressibility factor is defined as

For a van der Waals gas, substituting the critical constants $P_c=\frac{a}{27b^2}$, $V_c=3b$, and $T_c=\frac{8a}{27Rb}$ gives

For most real gases, the experimentally observed value is approximately

Step 1: Identify the Arrhenius two-temperature relation

The two-temperature form of the Arrhenius equation is

where $T_1=25+273=298\text{K}$ and $T_2=75+273=348\text{K}$.

Step 2: Substitute the known values

**Step 3: Solve for $k_2/k_1$

Result:

Step 1: Identify the process

For an isothermal process, the temperature $T$ remains constant throughout the expansion.

Step 2: Apply the work formula

The work done by the gas during isothermal expansion is given by:

where $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the absolute temperature, $V_f$ is the final volume, and $V_i$ is the initial volume.

Step 3: Determine the sign

Since the gas expands, we have $V_f>V_i$, which means:

Therefore, the work done by the gas is positive:

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution:

where $P$ is the partial pressure, $x$ is the mole fraction of the dissolved gas, and $K_h$ is the Henry's law constant.

For ${\text{CO}}_2$ dissolved in water at $25^{\circ }\text{C}$, the Henry's law constant is approximately:

**Step 1: Write the relationship between $\Delta G^{\circ }$ and $E_{\text{cell}}^{\circ }$

The standard Gibbs free energy change is related to the standard cell potential by:

where $n$ is the number of moles of electrons transferred, and $F=96500{\text{C mol}}^{-1}$ is the Faraday constant.

Step 2: Substitute the known values

For $n=2$ (typical two-electron transfer), $F=96500{\text{C mol}}^{-1}$, and $E_{\text{cell}}^{\circ }=0.34\text{V}$:

Result:

**Step 1: Determine $\Delta n$

For the reaction $2{\text{NO}}_2\rightleftharpoons {\text{N}}_2{\text{O}}_4$, the change in moles of gas is:

**Step 2: Apply the $K_p$–$K_c$ relationship**

The general relation is:

Substituting $\Delta n=-1$:

Le Chatelier's principle states that when a system at equilibrium is subjected to a change in pressure, the equilibrium shifts to partially oppose that change.

Step 1: Effect of increasing pressure

Increasing pressure favours the side of the equilibrium with fewer moles of gas. That is, for a general reaction

if $a+b>c+d$, increasing pressure shifts the equilibrium to the right (forward direction).

Step 2: Applying this to the specific reaction

For the reaction

the moles of gas on each side are:

Since $\Delta n_g<0$, increasing pressure favours the forward reaction, producing more ${\text{N}}_2{\text{O}}_4$.

Step 3: Evaluating the statement

The claim that *"increasing pressure always favours the forward reaction"* is not universally correct. It is only correct when the forward reaction proceeds in the direction of fewer moles of gas. A statement made without specifying the reaction, or applied to a reaction where $\Delta n_g=0$ (e.g., ${\text{H}}_2+{\text{I}}_2\rightleftharpoons 2\text{HI}$), would be incorrect, since

Conclusion

The statement that increasing pressure favours the forward reaction is only conditionally true — specifically when $\Delta n_g<0$. Without this qualification, the statement is:

Step 1: Identify the cathode and anode

In the Zn–Cu cell, copper undergoes reduction and zinc undergoes oxidation, so:

Step 2: Apply the cell potential formula

Result:

Step 1: Identify the known values

The two rate constants and corresponding temperatures are:

Step 2: Write the two-temperature Arrhenius equation

Step 3: Substitute the known values

Step 4: Evaluate the terms

**Step 5: Solve for $E_a$

Result:

For a general reaction, the half-life $t_{1/2}$ depends on the initial concentration $[A{]}_0$ according to the reaction order:

where $n$ is the order of reaction.

**Step 1: Zero-order reaction ($n=0$)**

The half-life *increases* with initial concentration, so $t_{1/2}$ is not independent of $[A{]}_0$.

**Step 2: First-order reaction ($n=1$)**

The half-life contains no $[A{]}_0$ term, so $t_{1/2}$ is independent of initial concentration.

**Step 3: Second-order reaction ($n=2$)**

The half-life *decreases* with initial concentration, so $t_{1/2}$ is not independent of $[A{]}_0$.

Conclusion:

Only for a first-order reaction is $t_{1/2}$ independent of the initial concentration $[A{]}_0$. The order of the reaction $2A\to B+C$ is therefore:

**Step 1: Identify the relationship between $\Delta G°$ and $K_c$

The standard Gibbs free energy change is related to the equilibrium constant by:

Step 2: Substitute the known values

Given $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$, $T=298\text{K}$, and $K_c=4$:

**Step 3: Evaluate $\ln (4)$

**Step 4: Calculate $\Delta G°$

Result:

Step 1: Recall the formula for molar conductivity

The molar conductivity ${\Lambda }_m$ is related to the specific conductance $\kappa$ and the molar concentration $C$ by:

where $\kappa$ is in $\text{S/cm}$, $C$ is in $\text{mol/L}$, and ${\Lambda }_m$ is in ${\text{S⋅cm}}^2\text{/mol}$.

Step 2: Substitute the given values

Given $\kappa =0.0141\text{S/cm}$ and $C=0.01\text{mol/L}$:

Step 3: State the final answer

Step 1: Identify the Henderson–Hasselbalch equation

The pH of a buffer is given by:

**Step 2: Calculate $p{K}_a$

Step 3: Substitute the concentrations

With $[{\text{A}}^{-}]=0.1\text{M}$ and $[\text{HA}]=0.1\text{M}$:

Result:

Step 1: Recall the entropy change formula for an ideal gas at constant temperature

For an isothermal process involving an ideal gas, the entropy change is given by

Step 2: Substitute the given values

For one mole of gas ($n=1$) with the volume doubled ($V_2=2V,V_1=V$),

Result

The entropy change for one mole of an ideal gas whose volume is doubled at constant temperature is

Step 1: Write the integrated rate law for a first-order reaction.

For a first-order reaction, the concentration at time $t$ is given by

Step 2: Substitute the known values.

With $k=0.05{\text{s}}^{-1}$ and $t=50\text{s}$:

Step 3: Evaluate the exponential.

Therefore:

Step 4: Express as a percentage.

Result:

The percentage of reactant remaining after $50\text{s}$ is

Step 1: Recall the Gibbs free energy equation

The Gibbs free energy change is given by

A reaction is spontaneous when $\Delta G<0$.

**Step 2: Analyze the signs of $\Delta H$ and $\Delta S$

Consider the case where $\Delta H<0$ (exothermic) and $\Delta S>0$ (entropy increases). Substituting these signs into the equation:

Since $T>0$ (temperature in Kelvin is always positive), the term $T\Delta S$ is strictly positive. Therefore:

Step 3: Conclusion

When $\Delta H<0$ and $\Delta S>0$, the Gibbs free energy is always negative regardless of temperature, meaning the reaction is spontaneous at all temperatures.

In an electrolytic cell, the anode (where oxidation occurs) is connected to the positive terminal of the external power source. In galvanic cells, the cathode is positive and the anode is negative. These relationships can be summarized as:

In both cell types, the anode is defined as the electrode at which oxidation occurs, and the cathode is defined as the electrode at which reduction occurs:

Step 1: Write the Gibbs free energy equation

The fundamental relationship is:

Step 2: Convert units

The entropy must be expressed in $\text{kJ/(mol⋅K)}$ to match $\Delta H$:

Step 3: Substitute the values

Given $T=300\text{K}$, $\Delta H=50\text{kJ/mol}$, and $\Delta S=0.100\text{kJ/(mol⋅K)}$:

Step 4: Interpret the result

Since $\Delta G>0$, the reaction is non-spontaneous at $T=300\text{K}$.

Step 1: Recall the half-life formula for a first-order reaction.

For a first-order reaction, the half-life is given by

Step 2: Substitute the given rate constant.

With $k=0.693{\text{s}}^{-1}$,

Step 3: State the final result.

The half-life of a first-order reaction is independent of the initial concentration.

**Step 1: Write the expression for $[{\text{H}}^{+}]$ in a weak acid solution**

For a weak acid, the hydrogen ion concentration is approximated by:

Step 2: Substitute the known values

Step 3: Calculate the pH