JEE Chemistry - MCQ Practice Questions

Step 1: Calculate the pOH

Step 2: Calculate the pH

Step 3: Classify the solution

Since $\text{pH}=6<7$, the solution is acidic.

Step 1: Identify the rate law

The given rate law is:

Step 2: Determine the order with respect to each reactant

Step 3: Calculate the overall order

The overall reaction order is the sum of all the exponents in the rate law:

This is a third-order reaction.

> Note: The reaction order is determined experimentally and need not match the stoichiometric coefficients. Here, the stoichiometric coefficient of $A$ is $2$ and of $B$ is $1$, which happen to equal the exponents in the rate law, but this is not always the case.

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

The correct relationship is:

Step 1: Identify the equation.

The standard Gibbs free energy change $\Delta G^{\circ }$ is related to the equilibrium constant $K$ by:

where $R$ is the universal gas constant and $T$ is the absolute temperature in Kelvin.

**Step 2: Interpret the sign of $\Delta G^{\circ }$.**

When $K>1$, we have $\ln (K)>0$, and therefore:

Step 3: State the key result.

This fundamental equation directly links thermodynamics and equilibrium:

Step 1: Identify the formula for isothermal expansion

For an isothermal expansion of an ideal gas, the entropy change is given by

Step 2: Substitute the known values

With $n=1\text{mol}$, $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$, $V_1=2\text{L}$, and $V_2=8\text{L}$:

Result

For the dissolution of AgCl, the equilibrium expression is

Step 1: Set up the expression in terms of solubility.

Let $s$ be the molar solubility of AgCl. Since each formula unit dissolves to give one ${\text{Ag}}^{+}$ ion and one ${\text{Cl}}^{-}$ ion, we have $[{\text{Ag}}^{+}]=[{\text{Cl}}^{-}]=s$, so

**Step 2: Solve for $s$.**

Result:

Step 1: Calculate moles of each reactant

Step 2: Determine moles after neutralization reaction

The NaOH neutralizes an equal amount of acetic acid:

**Step 3: Compute $\text{p}{K}_a$

Step 4: Apply the Henderson–Hasselbalch equation

Since the volume is common to both species, the ratio of moles equals the ratio of concentrations:

The coordination number is the number of ligands attached to the central metal ion.

Step 1: Identify the central metal ion and the ligands present.

In $[{\text{Cu(NH}}_3{)}_4{]}^{2+}$, the central metal ion is ${\text{Cu}}^{2+}$, and the ligands are ${\text{NH}}_3$ molecules.

Step 2: Count the number of ligands bonded to the central metal ion.

There are $4$ ammonia (${\text{NH}}_3$) ligands directly bonded to ${\text{Cu}}^{2+}$, so the coordination number is:

Step 1: Identify the integrated rate law for a second-order reaction.

For a reaction that is second-order with respect to reactant $A$, the integrated rate law is:

where $k$ is the rate constant, $t$ is time, and $[A{]}_0$ is the initial concentration of $A$.

Step 2: Eliminate incorrect options.

This is not the integrated rate law, so Option B is incorrect for this context.

Since this applies to a first-order reaction, Option D is also incorrect.

Step 3: Identify the correct answer.

The expression that correctly represents the second-order integrated rate law is:

Step 1: Recall the Gibbs free energy equation

The spontaneity of a reaction is determined by:

Step 2: Identify the given conditions

The reaction is exothermic, so:

Step 3: Analyze the condition for non-spontaneity

For the reaction to be non-spontaneous, we require $\Delta G>0$:

**Step 4: Determine the sign of $\Delta S$

Since $\Delta H<0$, the term $-T\Delta S$ must be positive and large enough to dominate at high $T$. For $-T\Delta S>0$ with $T>0$, we need:

Step 5: Verify the behavior at high temperature

As $T\to \infty$, the positive term $T|\Delta S|$ dominates, so $\Delta G>0$, confirming non-spontaneity at high temperatures.

Conclusion

For the exothermic reaction to become non-spontaneous at high temperatures, it must be true that:

Step 1: Identify the cathode and anode

In a galvanic cell, the electrode with the higher (more positive) standard reduction potential undergoes reduction at the cathode, while the electrode with the lower standard reduction potential undergoes oxidation at the anode.

Step 2: Apply the standard cell potential formula

Step 3: Substitute and calculate

Result:

Setting up the ICE table

Let $x$ be the partial pressure of ${\text{N}}_2{\text{O}}_4$ that dissociates (in atm).

For the equilibrium

the ICE table gives:

**Step 1: Write the $K_p$ expression**

Step 2: Expand and rearrange

Step 3: Solve the quadratic

Applying the quadratic formula to $4x^2+0.15x-0.15=0$:

Taking the positive root:

**Step 4: Calculate $P({\text{NO}}_2)$

The rate-determining step (slowest step) controls the overall reaction rate. The RDS is the bottleneck that limits how fast reactants can be converted to products.

Step 1: Define the rate-determining step.

For a multi-step mechanism, the overall rate law is governed by the slowest elementary step. If the mechanism has steps with rate constants $k_1,k_2,\dots ,k_n$, then the step with the smallest rate constant determines the overall rate:

Step 2: Write the rate law from the RDS.

Suppose the slow step is:

Then the overall rate is determined solely by this step:

Step 3: Confirm what the RDS controls.

Because every molecule must pass through the slow step, the overall rate cannot exceed the rate of the RDS, regardless of how fast the other steps are:

Any step after the RDS does not affect the overall rate, and any fast equilibrium step before the RDS contributes only through the steady-state or pre-equilibrium concentrations of intermediates.

Conclusion:

Step 1: Recall the relationship between activation energies and enthalpy

For any reaction, the forward and reverse activation energies are related by:

Step 2: Identify the given values

Step 3: Substitute and solve

Interpretation: Because the forward reaction is exothermic ($\Delta H<0$), the reverse reaction is endothermic. The reverse reaction must climb a larger energy barrier, since it must overcome both the original forward activation energy and the energy released by the forward reaction.

Final Answer:

Step 1: Write the Arrhenius two-temperature equation

The relationship between the rate constants at two temperatures is given by the Arrhenius equation in the form:

Step 2: Identify the known values

**Step 3: Substitute and solve for $E_a$

Step 4: Final Answer

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 $k=1.4\times 10^{-3}{\text{min}}^{-1}$,

Note: For first-order reactions, $t_{1/2}$ is independent of the initial concentration of the reactant.

Step 1: Identify the Rate-Determining Step

The rate-determining step is the slowest step in the mechanism. Here, Step 1 is slow:

Step 2: Write the Rate Law from the Rate-Determining Step

The rate law is determined solely by the elementary step that is slowest. For the bimolecular elementary step $A+B\to C$, the rate law is:

Step 3: Check for Intermediates

The species $C$ is a reaction intermediate (it is produced in Step 1 and consumed in Step 2). Since $C$ does not appear as a reactant in the rate-determining step, no further substitution or elimination is required.

The fast step,

does not influence the overall rate law because it follows the bottleneck step and proceeds much more quickly than Step 1.

Result:

The overall rate law is: