Chemical Thermodynamics – NEET UG ChemistryPractice Questions & PYQs

The first law of thermodynamics relates heat, internal energy, and work through the expression $q = \Delta E + W$. Because the container is insulated, there is no heat exchange, meaning $q=0$, while the mechanical paddle stirring performs work on the system, resulting \in $W \neq 0$. This energy transfer causes a temperature increase, confirming that the internal energy changes such that $\Delta E \neq 0$, which leads to the conclusion that $\Delta E = W \neq 0$.

The total enthalpy change of a chemical reaction is calculated by taking the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants. For the decomposition of liquid hydrogen peroxide, the products are liquid water and oxygen gas, while the reactant is liquid hydrogen peroxide. Since the standard enthalpy of formation for oxygen in its elemental state is zero, the calculation focuses on the provided values for water and hydrogen peroxide. Applying the expression $\Delta H^{\circ} =$[2 \Delta H_f^{\circ}(\text{H}_2\text{O}) + \Delta H_f^{\circ}(\text{O}_2)]$ - [2 \Delta H_f^{\circ}(\text{H}_2\text{O}_2)]$, we substitute the provided constants to compute$2(-286) + 0 - 2(-188)$. Simplifying the arithmetic to$-572 + 376$ leads to -196 kJ/mol.

We can determine the relationship between the enthalpies of combustion using Hess's Law, which states that the enthalpy change of a reaction is constant regardless of the pathway taken. The combustion of methane is represented as $\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$ with an enthalpy change of $-x$, while the combustion of methanol is $\mathrm{CH}_{3} \mathrm{OH}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$ with an enthalpy change of $-y$. Subtracting the methanol combustion equation from the methane combustion equation yields the target process $\mathrm{CH}_{4}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{OH}$. Since this target reaction is exothermic, its enthalpy change, which corresponds to $-x - (-y)$, must be a negative value, meaning $y - x < 0$ and therefore $x > y$.

Spontaneity in chemical processes is dictated by the sign of Gibbs free energy, where a negative value indicates a forward process occurs naturally, while a positive value implies that the reverse process is favored. Because the reduction $\mathrm{PbO}_2 \rightarrow \mathrm{PbO}$ has a $\Delta G_{298} < 0$, lead spontaneously shifts from the $+4$ oxidation state to the more stable $+2$ state. Conversely, since the reduction $\mathrm{SnO}_2 \rightarrow \mathrm{SnO}$ results in a $\Delta G_{298} > 0$, the reaction is non-spontaneous, establishing that $\mathrm{Sn}^{4+}$ is the more stable form relative to $\mathrm{Sn}^{2+}$.

The work done by or on a gas is determined by its volume change, defined as $w = -P\Delta V$. Because this process occurs at a constant volume, the change in volume $\Delta V$ is zero, which means no work is performed, so $w = 0$. According to the first law of thermodynamics, the internal energy change is given by $\Delta E = q + w$. Since the heat supplied $q$ is 500 J and there is no work done, the equation simplifies to $\Delta E = q + 0$, which means the internal energy change $\Delta E$ is 500 J.

The principle of additive properties for state functions like Gibbs Free Energy allows us to determine the energy change of a target chemical equation by combining known reactions. Because Gibbs Free Energy is a state function, the overall energy change of a reaction can be determined by algebraically summing the energy changes of individual step reactions, provided the equations are properly oriented to match the target.

To derive the target reaction $2 \mathrm{ZnS} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{ZnO} + 2 \mathrm{SO}_{2}$, we adjust the provided thermochemical equations. We reverse the reaction $2 \mathrm{Zn} + \mathrm{S}_{2} \rightarrow 2 \mathrm{ZnS}$ with $\Delta G^{\circ} = -293 \mathrm{~J}$ to obtain $2 \mathrm{ZnS} \rightarrow 2 \mathrm{Zn} + \mathrm{S}_{2}$, which inverts the sign of its free energy to $+293 \mathrm{~J}$. We then retain the equation $2 \mathrm{Zn} + \mathrm{O}_{2} \rightarrow 2 \mathrm{ZnO}$ with its value of $-616 \mathrm{~J}$ and the equation $\mathrm{S}_{2} + 2 \mathrm{O}_{2} \rightarrow 2 \mathrm{SO}_{2}$ with its value of $-408 \mathrm{~J}$. Adding these three balanced equations cancels out the intermediate species $2 \mathrm{Zn}$ and $\mathrm{S}_{2}$ from the reaction path, leaving the desired equation. Calculating the total free energy change involves summing the adjusted values, $293 - 616 - 408$, which results in $-731 \mathrm{~J}$.

Spontaneity in a chemical reaction is defined by a decrease in standard Gibbs free energy, meaning the system must have a negative $\Delta G^{\circ}$ value to proceed naturally without external intervention. This thermodynamic requirement is intrinsically linked to the standard cell potential through the equation $\Delta G^{\circ} = -n F E_{\text {cell}}^{\circ}$, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant. Since both $n$ and $F$ are always positive, the reaction can only occur spontaneously when the potential difference is positive, ensuring that the resulting value of $\Delta G^{\circ}$ is negative.

The entropy change during a phase transition is calculated by taking the quotient of the latent heat and the absolute temperature. Converting the given temperature from Celsius to Kelvin involves adding 273 to 27°C, which results in 300 K. With the formula $\Delta S = \frac{Q}{T}$, we can substitute the latent heat of 2930 J/mol and the temperature of 300 K into the expression. Dividing these values provides the final entropy change of 9.77 J/mol K.