NEET Chemistry - MCQ Practice Questions

Understanding:

We need the enthalpy change for the formation of 2 mol of ${\text{NH}}_3$ from its elements.

Formula:

Step 1: Apply Hess's Law

The standard enthalpy of formation of elements in their standard state is zero, so:

Answer:

The enthalpy change for the given reaction is $-92{\text{ kJ mol}}^{-1}$.

Quick Tip:

The standard enthalpy of formation is always per mole of product formed. When 2 moles are produced, multiply by 2. A common error is to use $-46$ directly without scaling.

Understanding:

We need $\Delta H_f^{\circ }$ for ${\text{CH}}_4(g)$, i.e., the enthalpy change for:

Formula:

By Hess's Law:

Step 1: Construct the target equation

Add equation 1 and $2\times$ equation 2, then subtract equation 3:

Step 2: Verify the equation

Using $\Delta H_{\text{rxn}}=\sum \Delta H_f^{\circ }(\text{products})-\sum \Delta H_f^{\circ }(\text{reactants})$ for combustion:

Answer:

The standard enthalpy of formation of ${\text{CH}}_4(g)$ is $-75{\text{ kJ mol}}^{-1}$.

Quick Tip:

To find $\Delta H_f^{\circ }$ of a compound using combustion data, rearrange the combustion equation algebraically. Reversing equation 3 changes the sign from $-890$ to $+890$.

Understanding:

We need $\Delta S_{\text{vap}}$ for vaporisation of water at its boiling point.

Formula:

At equilibrium (constant $T$ and $P$), $\Delta G=0$, so:

Step 1: Substitute values

Answer:

The entropy change of vaporisation is approximately $0.109{\text{ kJ mol}}^{-1}{\text{K}}^{-1}$.

Quick Tip:

Trouton's rule states that $\Delta S_{\text{vap}}\approx 88{\text{ J mol}}^{-1}{\text{K}}^{-1}$ for most non-associated liquids. Water's value ($\approx 109{\text{ J mol}}^{-1}{\text{K}}^{-1}$) is higher due to hydrogen bonding.

Understanding:

We need the temperature above which $\Delta G>0$ (reaction becomes non-spontaneous).

Formula:

The crossover temperature (where $\Delta G=0$) is:

Step 1: Identify the spontaneity pattern

Since both $\Delta H<0$ and $\Delta S<0$, the $-T\Delta S$ term is positive (adding to $\Delta G$). The reaction is spontaneous at low $T$ but becomes non-spontaneous at high $T$.

Step 2: Find the crossover temperature

Step 3: Interpret

Below $2500\text{ K}$: $\Delta G<0$ (spontaneous). Above $2500\text{ K}$: $\Delta G>0$ (non-spontaneous).

Answer:

The reaction becomes non-spontaneous above approximately $2500\text{ K}$.

Quick Tip:

When $\Delta H$ and $\Delta S$ have the same sign, there is always a crossover temperature. Always convert $\Delta H$ to joules (or $\Delta S$ to kJ) before dividing to keep units consistent.

Understanding:

We use bond enthalpies to find $\Delta H$ for the reaction.

Formula:

Step 1: Bonds broken (reactants)

Step 2: Bonds formed (products)

Two moles of $\text{HCl}$ are formed, each with one $\text{H–Cl}$ bond:

Step 3: Calculate $\Delta H$

Answer:

The enthalpy change for the reaction is $-184{\text{ kJ mol}}^{-1}$.

Quick Tip:

Bond breaking is always endothermic (positive contribution) and bond forming is always exothermic (negative contribution). A common error is reversing this sign convention.

Understanding:

We want $\Delta H_f^{\circ }$ for ${\text{C}}_2{\text{H}}_6(g)$, i.e., $\Delta H$ for:

Formula:

By Hess's Law:

Step 1: Apply the formula

Answer:

The standard enthalpy of formation of ethane is $-84{\text{ kJ mol}}^{-1}$.

Quick Tip:

When using heats of combustion to find $\Delta H_f^{\circ }$, the formula involves subtracting the compound's combustion enthalpy (not adding it). This is the reverse of using formation enthalpies directly.

Understanding:

We analyse $\Delta U$, $q$, and $w$ for isothermal reversible expansion of an ideal gas.

Formula:

First Law of Thermodynamics:

For an ideal gas, internal energy depends only on temperature:

Step 1: Determine $\Delta U$

Isothermal means $\Delta T=0$. For an ideal gas:

Step 2: Determine $w$

During expansion, the gas does work on the surroundings, so work done ON the system is negative (IUPAC convention):

Step 3: Determine $q$ using the First Law

The gas absorbs heat from surroundings to do work.

Answer:

For an isothermal reversible expansion of an ideal gas: $\Delta U=0$, $q>0$, $w<0$.

Quick Tip:

For any isothermal process involving an ideal gas, $\Delta U=0$ is always the starting point, since $U$ is a function of temperature only.

Understanding:

We find the enthalpy of ionisation of $\text{HCN}$ from the enthalpy data.

Formula:

Neutralisation of a weak acid involves two steps:

Step 1: Write the two processes

For a strong acid: ${\text{H}}^{+}(aq)+{\text{OH}}^{-}(aq)\to {\text{H}}_2\text{O}(l)$, $\Delta H=-57{\text{ kJ mol}}^{-1}$.

For HCN, the extra step is ionisation: $\text{HCN}(aq)\to {\text{H}}^{+}(aq)+{\text{CN}}^{-}(aq)$, $\Delta H_{\text{ion}}=?$

Step 2: Apply the relation

Answer:

The enthalpy of ionisation of $\text{HCN}$ is $+45{\text{ kJ mol}}^{-1}$.

Quick Tip:

The enthalpy of ionisation of a weak acid is always positive (endothermic), explaining why the heat evolved during its neutralisation is less than $57{\text{ kJ mol}}^{-1}$.

Understanding:

We analyse the thermodynamic quantities for adiabatic reversible compression of one mole of an ideal gas.

Formula:

First Law of Thermodynamics:

For an adiabatic process:

Step 1: Determine $q$

By definition, an adiabatic process has no heat exchange with surroundings:

Step 2: Determine $w$

Compression means the surroundings do work ON the gas, so $\Delta V<0$. Using the IUPAC sign convention ($w=-P_{\text{ext}}\Delta V$):

Step 3: Determine $\Delta U$

This also confirms that temperature rises during adiabatic compression, since $\Delta U=n{C}_v\Delta T>0$.

Answer:

For adiabatic reversible compression: $q=0$, $w>0$, and $\Delta U>0$.

Quick Tip:

In an adiabatic compression, all the work done on the gas stays inside as internal energy, raising the temperature. This is why a bicycle pump gets warm when you inflate a tyre quickly.

Understanding:

We find the enthalpy of hydration using the Born–Haber cycle relationship for dissolution.

Formula:

The enthalpy of solution is related to lattice and hydration enthalpies by:

Step 1: Solve for $\Delta H_{\text{hydration}}$

Step 2: Verify the sign

Hydration is always exothermic (water molecules stabilise the ions), so $\Delta H_{\text{hydration}}<0$. The result $-784{\text{ kJ mol}}^{-1}$ is consistent.

Answer:

The enthalpy of hydration of $\text{NaCl}$ is $-784{\text{ kJ mol}}^{-1}$.

Quick Tip:

Hydration enthalpy is always negative. If the magnitude of hydration enthalpy exceeds the lattice enthalpy, the salt dissolves with a negative $\Delta H_{\text{solution}}$ (exothermic dissolution). For $\text{NaCl}$, they are nearly equal, giving a near-zero $\Delta H_{\text{solution}}$.