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Q5.21:
Comment on the thermodynamic stability of (\text{NO(g)}), given
(\frac{1}{2} \text{N}_{2}(\text{g}) + \frac{1}{2}\text{O}_{2}(\text{g}) \rightarrow \text{NO(g)}); (\Delta_{r}H^{\ominus} = 90 \text{ kJ mol}^{-1})
(\text{NO(g)} + \frac{1}{2}\text{O}_{2}(\text{g}) \rightarrow \text{NO}_{2}(\text{g}): \Delta_{r}H^{\ominus}= –74 \text{ kJ mol}^{-1})
Solution :
Step 1: Evaluation of Thermodynamic Stability
Thermodynamic stability of a compound is determined by its enthalpy of formation ($\Delta_{f}H^{\ominus}$). A compound is considered thermodynamically stable relative to its constituent elements if its enthalpy of formation is negative (exothermic). If the enthalpy of formation is positive (endothermic), the compound is thermodynamically unstable relative to its constituent elements.
Step 2: Analysis of the Given Enthalpy Data
We are provided with the following thermochemical equations:
- Reaction (1): $\frac{1}{2} N_{2}(g) + \frac{1}{2} O_{2}(g) \rightarrow NO(g)$; $\Delta_{r}H^{\ominus} = +90 \text{ kJ mol}^{-1}$
- Reaction (2): $NO(g) + \frac{1}{2} O_{2}(g) \rightarrow NO_{2}(g)$; $\Delta_{r}H^{\ominus} = -74 \text{ kJ mol}^{-1}$
Step 3: Determining the Stability of NO(g)
The enthalpy of formation for $NO(g)$ is given directly by Reaction (1), as it represents the formation of one mole of $NO$ from its constituent elements in their standard states [Per the definition of standard enthalpy of formation, $\Delta_{f}H^{\ominus}$].
Since $\Delta_{r}H^{\ominus} = +90 \text{ kJ mol}^{-1} > 0$, the formation of $NO(g)$ is an endothermic process. This indicates that $NO(g)$ possesses higher energy than its constituent elements ($N_2$ and $O_2$). Therefore, $NO(g)$ is thermodynamically unstable with respect to its elements at standard conditions.
Step 4: Comparative Stability Context
While $NO(g)$ is unstable, we can observe its reactivity via Reaction (2). By applying Hess's Law of Constant Heat Summation [Per Hess's Law], we can determine the enthalpy of formation for $NO_2(g)$:
| Reaction | $\Delta_{r}H^{\ominus} (\text{kJ mol}^{-1})$ |
|---|---|
| $\frac{1}{2}N_2 + \frac{1}{2}O_2 \rightarrow NO$ | $+90$ |
| $NO + \frac{1}{2}O_2 \rightarrow NO_2$ | $-74$ |
| Net: $\frac{1}{2}N_2 + O_2 \rightarrow NO_2$ | $+16$ |
The calculation reveals that $NO(g)$ tends to react spontaneously with oxygen to form $NO_2(g)$ because the process releases energy ($\Delta H = -74 \text{ kJ mol}^{-1}$), despite $NO_2$ itself remaining endothermic relative to its elements.
Step 5: Visual Representation of Energy Levels
Final Solution: The positive enthalpy of formation ($\Delta_{f}H^{\ominus} = +90 \text{ kJ mol}^{-1}$) confirms that $NO(g)$ is thermodynamically unstable with respect to its constituent elements at standard conditions.
More Questions from Class 11 Physics Thermodynamics EXERCISES
- Q5.1: Choose the correct answer. A thermodynamic state function is a quantity
- Q5.10: Calculate the enthalpy change on freezing of 1.0 mol of water at (10.0^{\circ}\text{C}) to ice at (–10.0^{\circ}\text{C}). (\Delta_{\text{fus}}H = 6.03 \text{ kJ mol}^{-1}) at (0^{\circ}\text{C}). (C_{p} [\text{H}_{2}\text{O(l)}] = 75.3 \text{ J mol}^{-1} \text{ K}^{-1}) (C_{p} [\text{H}_{2}\text{O(s)}] = 36.8 \text{ J mol}^{-1} \text{ K}^{-1})
- Q5.11: Enthalpy of combustion of carbon to (\text{CO}_{2}) is (–393.5 \text{ kJ mol}^{-1}). Calculate the heat released upon formation of 35.2 g of (\text{CO}_{2}) from carbon and dioxygen gas.
- Q5.12: Enthalpies of formation of (\text{CO(g)}), (\text{CO}_{2}(\text{g})), (\text{N}_{2}\text{O(g)}) and (\text{N}_{2}\text{O}_{4}(\text{g})) are –110, – 393, 81 and (9.7 \text{ kJ mol}^{-1}) respectively. Find the value of (\Delta_{r}H) for the reaction: (\text{N}_{2}\text{O}_{4}(\text{g}) + 3\text{CO(g)} \rightarrow \text{N}_{2}\text{O(g)} + 3\text{CO}_{2}(\text{g}))
- Q5.13: Given (\text{N}_{2}(\text{g}) + 3\text{H}_{2}(\text{g}) \rightarrow 2\text{NH}_{3}(\text{g}); \Delta_{r}H^{\ominus} = –92.4 \text{ kJ mol}^{-1}) What is the standard enthalpy of formation of (\text{NH}_{3}) gas?
- Q5.14: Calculate the standard enthalpy of formation of (\text{CH}_{3}\text{OH(l)}) from the following data: (\text{CH}_{3}\text{OH (l)} + \frac{3}{2} \text{O}_{2}(\text{g}) \rightarrow \text{CO}_{2}(\text{g}) + 2\text{H}_{2}\text{O(l)}) ; (\Delta_{r}H^{\ominus} = –726 \text{ kJ mol}^{-1}) (\text{C(graphite)} + \text{O}_{2}(\text{g}) \rightarrow \text{CO}_{2}(\text{g})) ; (\Delta_{c}H^{\ominus} = –393 \text{ kJ mol}^{-1}) (\text{H}_{2}(\text{g}) + \frac{1}{2} \text{O}_{2}(\text{g}) \rightarrow \text{H}_{2}\text{O(l)}); (\Delta_{f}H^{\ominus} = –286 \text{ kJ mol}^{-1}).
- Q5.15: Calculate the enthalpy change for the process (\text{CCl}_{4}(\text{g}) \rightarrow \text{C(g)} + 4 \text{Cl(g)}) and calculate bond enthalpy of C – Cl in (\text{CCl}_{4}(\text{g})). (\Delta_{\text{vap}}H^{\ominus}(\text{CCl}_{4}) = 30.5 \text{ kJ mol}^{-1}). (\Delta_{f}H^{\ominus} (\text{CCl}_{4}) = –135.5 \text{ kJ mol}^{-1}). (\Delta_{a}H^{\ominus} (\text{C}) = 715.0 \text{ kJ mol}^{-1}), where (\Delta_{a}H^{\ominus}) is enthalpy of atomisation (\Delta_{a}H^{\ominus} (\text{Cl}_{2}) = 242 \text{ kJ mol}^{-1})
- Q5.16: For an isolated system, (\Delta U = 0), what will be (\Delta S) ?
- Q5.17: For the reaction at 298 K, (2A + B \rightarrow C) (\Delta H = 400 \text{ kJ mol}^{-1}) and (\Delta S = 0.2 \text{ kJ K}^{-1} \text{ mol}^{-1}) At what temperature will the reaction become spontaneous considering (\Delta H) and (\Delta S) to be constant over the temperature range.
- Q5.18: For the reaction, (2 \text{Cl(g)} \rightarrow \text{Cl}_{2}(\text{g})), what are the signs of (\Delta H) and (\Delta S) ?
- Q5.19: For the reaction (2 \text{A(g)} + \text{B(g)} \rightarrow 2\text{D(g)}) (\Delta U^{\ominus} = –10.5 \text{ kJ}) and (\Delta S^{\ominus} = –44.1 \text{ JK}^{-1}). Calculate (\Delta G^{\ominus}) for the reaction, and predict whether the reaction may occur spontaneously.
- Q5.2: For the process to occur under adiabatic conditions, the correct condition is:
- Q5.20: The equilibrium constant for a reaction is 10. What will be the value of (\Delta G^{\ominus}) ? (R = 8.314 \text{ JK}^{-1} \text{ mol}^{-1}), (T = 300 \text{ K}).
- Q5.22: Calculate the entropy change in surroundings when 1.00 mol of (\text{H}_{2}\text{O(l)}) is formed under standard conditions. (\Delta_{f}H^{\ominus} = –286 \text{ kJ mol}^{-1}).
- Q5.3: The enthalpies of all elements in their standard states are:
- Q5.4: (\Delta U^{\ominus}) of combustion of methane is (– X \text{ kJ mol}^{-1}). The value of (\Delta H^{\ominus}) is
- Q5.5: The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, (–890.3 \text{ kJ mol}^{-1}), (–393.5 \text{ kJ mol}^{-1}), and (–285.8 \text{ kJ mol}^{-1}) respectively. Enthalpy of formation of (\text{CH}_{4}(\text{g})) will be
- Q5.6: A reaction, (A + B \rightarrow C + D + q) is found to have a positive entropy change. The reaction will be
- Q5.7: In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
- Q5.8: The reaction of cyanamide, (\text{NH}_{2}\text{CN (s)}), with dioxygen was carried out in a bomb calorimeter, and (\Delta U) was found to be (–742.7 \text{ kJ mol}^{-1}) at 298 K. Calculate enthalpy change for the reaction at 298 K. (\text{NH}_{2}\text{CN(g)} + \frac{3}{2}\text{O}_{2}(\text{g}) \rightarrow \text{N}_{2}(\text{g}) + \text{CO}_{2}(\text{g}) + \text{H}_{2}\text{O(l)})
- Q5.9: Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from (35^{\circ}\text{C}) to (55^{\circ}\text{C}). Molar heat capacity of Al is (24 \text{ J mol}^{-1} \text{ K}^{-1}).
- Q12.1: A geyser heats water flowing at the rate of 3.0 litres per minute from $27^\circ C$ to $77^\circ C$. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 \times 10^4 J/g$ ?
- Q12.10: A refrigerator is to maintain eatables kept inside at $9^\circ C$. If room temperature is $36^\circ C$, calculate the coefficient of performance.
- Q12.2: What amount of heat must be supplied to $2.0 \times 10^{-2}$ kg of nitrogen (at room temperature) to raise its temperature by $45^\circ C$ at constant pressure ? (Molecular mass of $N_2 = 28$; $R = 8.3 J mol^{-1} K^{-1}$.)
- Q12.3(a): Explain why (a) Two bodies at different temperatures $T_1$ and $T_2$ if brought in thermal contact do not necessarily settle to the mean temperature $(T_1 + T_2)/2$.
- Q12.3(b): Explain why (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
- Q12.3(c): Explain why (c) Air pressure in a car tyre increases during driving.
- Q12.3(d): Explain why (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
- Q12.4: A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?
- Q12.5: In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J)
- Q12.6(a): Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B ?
- Q12.6(b): Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (b) What is the change in internal energy of the gas ?
- Q12.6(c): Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (c) What is the change in the temperature of the gas ?
- Q12.6(d): Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?
- Q12.7: A steam engine delivers $5.4 \times 10^8 J$ of work per minute and services $3.6 \times 10^9 J$ of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
- Q12.8: An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
- Q12.9: A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13). Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
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