This sheet covers the key Thermodynamics Class 11 Formulas from NCERT Chemistry Part I, Chapter 5. It includes the first law of thermodynamics, work, enthalpy, heat capacity, entropy, Gibbs energy, and their relationships. Each formula is grouped by topic, with symbol meanings, when-to-use guidance, and original worked examples.
For the detailed explanations, derivations, and sign conventions, refer to the Class 11 Chemistry Formulas page. The NCERT solutions for this chapter can be found on the Chemistry Formulas page.
Formulas at a Glance
| Purpose (what you are finding) | Formula |
|---|---|
| Change in internal energy (first law) | \(\Delta U = q + w\) |
| Work done by/on system (constant external pressure) | \(w = -p_{\text{ex}} \Delta V\) |
| Reversible isothermal work for ideal gas | \(w_{\text{rev}} = -nRT\ln\frac{V_f}{V_i} = -2.303\,nRT\log\frac{V_f}{V_i}\) |
| Heat at constant volume | \(q_V = \Delta U\) |
| Heat at constant pressure (enthalpy change) | \(q_p = \Delta H\) |
| Enthalpy definition | \(H = U + pV\) |
| Relation between ΔH and ΔU | \(\Delta H = \Delta U + \Delta n_g RT\) |
| Heat capacity | \(q = C\Delta T = c\,m\,\Delta T\) |
| Difference between Cp and Cv (ideal gas) | \(C_p – C_v = R\) |
| Standard enthalpy of reaction | \(\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) – \sum b_i \Delta_f H^\ominus(\text{reactants})\) |
| Hess’s law (constant heat summation) | \(\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \dots\) |
| Bond enthalpy approximation (gas phase) | \(\Delta_r H^\ominus \approx \sum \text{bond enthalpies}_{\text{reactants}} – \sum \text{bond enthalpies}_{\text{products}}\) |
| Entropy change (reversible process) | \(\Delta S = \dfrac{q_{\text{rev}}}{T}\) |
| Total entropy change for spontaneity | \(\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \gt 0\) |
| Entropy change of surroundings (constant p) | \(\Delta S_{\text{surr}} = -\dfrac{\Delta H_{\text{sys}}}{T}\) |
| Gibbs energy definition | \(G = H – TS\) |
| Gibbs energy change (constant T) | \(\Delta G = \Delta H – T\Delta S\) |
| Gibbs energy and equilibrium constant | \(\Delta_r G^\ominus = -RT\ln K = -2.303\,RT\log K\) |
All Formulas, Grouped by Topic
First Law of Thermodynamics
The mathematical statement of the first law is:
\[\Delta U = q + w\]
For an isolated system, \(q = 0\) and \(w = 0\), so \(\Delta U = 0\).
Pressure-Volume Work
When a system expands or contracts against a constant external pressure:
\[w = -p_{\text{ex}} \Delta V\]
For reversible processes, the external pressure is infinitesimally close to the internal pressure of the gas. For an ideal gas undergoing isothermal reversible expansion or compression:
\[w_{\text{rev}} = -\int_{V_i}^{V_f} p\,dV = -nRT \ln\frac{V_f}{V_i} = -2.303\,nRT\log\frac{V_f}{V_i}\]
In free expansion (vacuum), \(p_{\text{ex}} = 0\), so \(w = 0\).
Heat at Constant Volume and Constant Pressure
At constant volume, no work is done:
\[q_V = \Delta U\]
At constant pressure, the heat absorbed equals the change in enthalpy:
\[q_p = \Delta H\]
Enthalpy
Enthalpy is defined as:
\[H = U + pV\]
For a finite change at constant pressure:
\[\Delta H = \Delta U + p\Delta V\]
Using the ideal gas law, the relation between \(\Delta H\) and \(\Delta U\) for reactions involving gases is:
\[\Delta H = \Delta U + \Delta n_g RT\]
where \(\Delta n_g\) = (moles of gaseous products) – (moles of gaseous reactants).
Heat Capacity
The heat required to raise the temperature of a substance:
\[q = C\Delta T = c\,m\,\Delta T\]
For an ideal gas, the molar heat capacities at constant pressure and constant volume are related by:
\[C_p – C_v = R\]
Reaction Enthalpy and Standard Enthalpy of Formation
The standard enthalpy change for a reaction is calculated from standard enthalpies of formation:
\[\Delta_r H^\ominus = \sum\limits_i a_i \Delta_f H^\ominus(\text{products}) – \sum\limits_i b_i \Delta_f H^\ominus(\text{reactants})\]
where \(a_i\) and \(b_i\) are the stoichiometric coefficients.
Hess’s Law
If a reaction occurs in several steps, the overall enthalpy change is the sum of the enthalpy changes of the individual steps:
\[\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \dots\]
Bond Enthalpies (Gas Phase Approximation)
The standard reaction enthalpy can be estimated from bond enthalpies when all reactants and products are gases:
\[\Delta_r H^\ominus \approx \sum \text{bond enthalpies}_{\text{reactants}} – \sum \text{bond enthalpies}_{\text{products}}\]
Entropy
For a reversible process, entropy change is defined as:
\[\Delta S = \frac{q_{\text{rev}}}{T}\]
The total entropy change for a process and its surroundings determines spontaneity.
For a spontaneous process:
\[\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \gt 0\]
At constant pressure, the entropy change of the surroundings is:
\[\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}\]
Gibbs Energy
Gibbs energy (or free energy) is defined as:
\[G = H – TS\]
At constant temperature:
\[\Delta G = \Delta H – T\Delta S\]
Spontaneity criterion at constant pressure and temperature:
- \(\Delta G \lt 0\) → spontaneous
- \(\Delta G \gt 0\) → non‑spontaneous
- \(\Delta G = 0\) → equilibrium
Gibbs Energy and Equilibrium Constant
The standard Gibbs energy change is related to the equilibrium constant:
\[\Delta_r G^\ominus = -RT\ln K = -2.303\,RT\log K\]
What Each Symbol Means
| Symbol | Meaning | Unit (SI) |
|---|---|---|
| \(U\) | Internal energy of the system | J |
| \(\Delta U\) | Change in internal energy | J |
| \(q\) | Heat transferred to the system | J |
| \(w\) | Work done on the system | J |
| \(p\) | Pressure | Pa (N m–2) |
| \(V\) | Volume | m3 |
| \(T\) | Absolute temperature | K |
| \(n\) | Amount of substance (number of moles) | mol |
| \(R\) | Universal gas constant | J K–1 mol–1 |
| \(H\) | Enthalpy of the system | J |
| \(\Delta H\) | Change in enthalpy | J |
| \(\Delta n_g\) | Change in number of moles of gaseous species | mol (dimensionless in ratio) |
| \(C\) | Heat capacity of the system | J K–1 |
| \(c\) | Specific heat capacity | J kg–1 K–1 |
| \(C_p\) | Molar heat capacity at constant pressure | J K–1 mol–1 |
| \(C_v\) | Molar heat capacity at constant volume | J K–1 mol–1 |
| \(\Delta_f H^\ominus\) | Standard molar enthalpy of formation | kJ mol–1 |
| \(\Delta_r H^\ominus\) | Standard enthalpy of reaction | kJ mol–1 |
| \(S\) | Entropy of the system | J K–1 |
| \(\Delta S\) | Change in entropy | J K–1 |
| \(G\) | Gibbs energy (free energy) | J |
| \(\Delta G\) | Change in Gibbs energy | J |
| \(K\) | Equilibrium constant (dimensionless) | — |
When to Use Each Formula
| Formula | When to use it |
|---|---|
| \(\Delta U = q + w\) | Always valid for any closed system. Use to find the change in internal energy when heat and work are known. |
| \(w = -p_{\text{ex}}\Delta V\) | When the system expands or contracts against a constant external pressure (irreversible process). |
| \(w_{\text{rev}} = -nRT\ln(V_f/V_i)\) | Only for isothermal reversible expansion/compression of an ideal gas. Condition: \(T\) constant, gas ideal. |
| \(q_V = \Delta U\) | When the reaction is carried out at constant volume (e.g., in a bomb calorimeter). |
| \(q_p = \Delta H\) | When the reaction occurs at constant pressure (e.g., open beaker, atmospheric pressure). |
| \(\Delta H = \Delta U + \Delta n_g RT\) | To convert between enthalpy change and internal energy change for reactions involving gases, when \(p\) and \(T\) are constant. |
| \(C_p – C_v = R\) | For an ideal gas only. Use to find one heat capacity if the other is known. |
| \(\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) – \sum b_i \Delta_f H^\ominus(\text{reactants})\) | To calculate the standard enthalpy of any reaction from standard enthalpies of formation of reactants and products. |
| Hess’s law: \(\Delta_r H = \sum \Delta H_{\text{steps}}\) | When the reaction cannot be measured directly; combine known enthalpy changes of intermediate reactions. |
| \(\Delta S = q_{\text{rev}}/T\) | To calculate the entropy change of a system for a reversible isothermal process. |
| \(\Delta S_{\text{surr}} = -\Delta H_{\text{sys}}/T\) | To find the entropy change of the surroundings when the system undergoes a process at constant pressure. |
| \(\Delta G = \Delta H – T\Delta S\) | To determine spontaneity at constant pressure and temperature. Use when both enthalpy and entropy changes are known. |
| \(\Delta_r G^\ominus = -RT\ln K\) | To relate the standard Gibbs energy change to the equilibrium constant at a given temperature. |
Worked Examples
Example 1: Applying the First Law
A gas absorbs 250 J of heat and does 80 J of work on the surroundings. Calculate the change in internal energy of the gas.
Step 1: Identify the signs.
Heat absorbed by system → \(q = +250\,\text{J}\).
Work done by system → \(w = -80\,\text{J}\).
Step 2: Use the first law: \(\Delta U = q + w\).
\[ \Delta U = 250\,\text{J} + (-80\,\text{J}) = 170\,\text{J} \]
Final answer: \(\Delta U = 170\,\text{J}\). The internal energy of the gas increases by 170 J.
Example 2: Reversible Isothermal Work
Calculate the work done when 2.0 mol of an ideal gas expands reversibly and isothermally at 300 K from 5.0 L to 15.0 L. (R = 8.314 J K⁻¹ mol⁻¹)
- Step 1: Use the formula for reversible isothermal work: \(w_{\text{rev}} = -nRT\ln\dfrac{V_f}{V_i}\).
- Step 2: Substitute values: \(n=2.0\) mol, \(R=8.314\,\text{J K}^{-1}\text{mol}^{-1}\), \(T=300\,\text{K}\), \(V_f=15.0\,\text{L}\), \(V_i=5.0\,\text{L}\).
Note: volume ratio is dimensionless, so units cancel.
\[ w_{\text{rev}} = -2.0 \times 8.314 \times 300 \times \ln\frac{15.0}{5.0} \]
\[ \ln 3 = 1.0986 \]
\[ w_{\text{rev}} = -2.0 \times 8.314 \times 300 \times 1.0986 \]
\[ w_{\text{rev}} = -5480\,\text{J} \]
Final answer: \(w = -5.48 \times 10^3\,\text{J}\). The negative sign indicates work is done by the system.
Example 3: Using ΔG = ΔH – TΔS
A reaction has \(\Delta H = +50\,\text{kJ mol}^{-1}\) and \(\Delta S = +120\,\text{J K}^{-1}\text{mol}^{-1}\) at 298 K. Is the reaction spontaneous at this temperature? At what temperature does it become spontaneous?
- Step 1: Convert \(\Delta S\) to kJ: \(120\,\text{J K}^{-1}\text{mol}^{-1} = 0.120\,\text{kJ K}^{-1}\text{mol}^{-1}\).
- Step 2: Calculate \(\Delta G\) at 298 K using \(\Delta G = \Delta H – T\Delta S\).
\[ \Delta G = 50 – (298 \times 0.120) = 50 – 35.76 = 14.24\,\text{kJ mol}^{-1} \]
- Step 1: Since \(\Delta G \gt 0\), the reaction is non‑spontaneous at 298 K.
- Step 2: The reaction becomes spontaneous when \(\Delta G \lt 0\).
Set \(\Delta G = 0\) and solve for \(T\):
\[ 0 = \Delta H – T\Delta S \quad \Rightarrow \quad T = \frac{\Delta H}{\Delta S} = \frac{50\,\text{kJ mol}^{-1}}{0.120\,\text{kJ K}^{-1}\text{mol}^{-1}} = 416.7\,\text{K} \]
Final answer: The reaction is non‑spontaneous at 298 K. It becomes spontaneous above \(\approx 417\,\text{K}\).
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using \(w = -p\Delta V\) for reversible processes without integrating. | For reversible processes, \(p_{\text{ex}} = p_{\text{int}}\); use \(w_{\text{rev}} = -\int p\,dV\). For an ideal gas isothermal process, use the logarithmic formula. | If the process is reversible and the pressure changes, check that you integrated or used the derived expression. |
| Confusing \(\Delta H\) with \(\Delta U\) in reactions involving gases. | Always use \(\Delta H = \Delta U + \Delta n_g RT\) when gases are present. If \(\Delta n_g = 0\), then \(\Delta H = \Delta U\). | Calculate \(\Delta n_g\) from the balanced equation; if it is non‑zero, the two values differ. |
| Forgetting that \(C_p – C_v = R\) applies only to ideal gases. | This relation is derived from the ideal gas law. For real gases, the difference is not exactly \(R\). | If the problem states “ideal gas”, the relation holds. Otherwise, do not assume it. |
| Using \(\Delta S = q/T\) for irreversible processes. | \(\Delta S = q_{\text{rev}}/T\) is defined for a reversible path. For an irreversible process, you must calculate \(\Delta S\) by considering a reversible alternative. | If the process is not reversible, you cannot use \(q_{\text{actual}}/T\); find a reversible path between the same states. |
| Misinterpreting sign of \(\Delta G\): thinking \(\Delta G \lt 0\) always means fast reaction. | \(\Delta G \lt 0\) only indicates that the reaction is thermodynamically spontaneous; it says nothing about the rate. | Spontaneity does not imply speed. A reaction can be spontaneous but extremely slow (e.g., diamond to graphite). |
Frequently Asked Questions
What is the difference between \(\Delta U\) and \(\Delta H\)?
\(\Delta U\) is the change in internal energy (heat at constant volume). \(\Delta H\) is the change in enthalpy (heat at constant pressure). They are related by \(\Delta H = \Delta U + \Delta n_g RT\). For reactions where the number of gas moles does not change, \(\Delta H = \Delta U\).
When is \(w = -p_{\text{ex}}\Delta V\) used?
This formula is used for irreversible expansion or compression against a constant external pressure. For reversible processes, the external pressure changes continuously, so we integrate.
How do I decide whether a reaction is spontaneous using \(\Delta G\)?
At constant pressure and temperature, if \(\Delta G \lt 0\) the reaction is spontaneous; if \(\Delta G \gt 0\) it is non‑spontaneous; if \(\Delta G = 0\) the system is at equilibrium. The sign of \(\Delta G\) depends on both \(\Delta H\) and \(\Delta S\) and the temperature.
What is the significance of \(\Delta_r G^\ominus = -RT\ln K\)?
This equation connects thermodynamics with equilibrium. A large negative \(\Delta_r G^\ominus\) gives a large \(K \gt 1\), meaning the reaction proceeds far towards products. A positive \(\Delta_r G^\ominus\) gives a small \(K \lt 1\), favouring reactants.
Reference: NCERT Class 11 Chemistry textbook, chapter Thermodynamics.
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