Enthalpy H = U + PV; ΔH for reactions at constant P

Overview

Enthalpy (H) is a state function that measures the total heat content of a system. It's defined as H = U + PV, where U is internal energy, P is pressure, and V is volume. For chemical reactions at constant pressure (most lab conditions), ΔH directly equals the heat exchanged with surroundings.

Why Do We Need Enthalpy?

WHY define H = U + PV? Because when P is constant, the changeΔH perfectly captures the heat exchanged, making thermochemistry experiments straightforward.

Derivation from First Principles

Formula Defining Enthalpy

Start with the First Law: $\Delta U = q + w$

Step 1: For PV work at constant pressure: $w = -P_{\text{ext}}\Delta V$

Why negative? When gas expands (ΔV > 0), the system does work on surroundings, losing energy.

Step 2: At constant pressure where $P_{\text{ext}} = P_{\text{system}}$ : $\Delta U = q_P - P\Delta V$

Rearrange for heat at constant pressure: $q_P = \Delta U + P\Delta V$

Step 3: Factor the right side: $q_P = (U_{\text{final}} - U_{\text{initial}}) + P(V_{\text{final}} - V_{\text{initial}})$ $q_P = (U_{\text{final}} + PV_{\text{final}}) - (U_{\text{initial}} + PV_{\text{initial}})$ Step 4: Define the state function: $H \equiv U + PV$ Then: $\boxed{\Delta H = q_P}$

At constant pressure, enthalpy change equals heat absorbed/released.

WHY this works: The PV term accounts for expansion work automatically. When you measure heat flow in a constant-pressure calorimeter, you're directly measuring ΔH.

Physical Meaning of Each Term

ΔH for Chemical Reactions

WHY this sign convention? When products have less enthalpy than reactants, the "missing" energy was released as heat to surroundings — exothermic.

Worked Examples

Worked example Example 1: Combustion of Methane

$\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H° = -890.3 \text{ kJ/mol}$

Given: Reaction occurs at 1 atm (constant P), 25°C Question: How much heat is released when 16 g of CH₄ burns? Solution: Step 1**: Convert mass to moles $n = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1.0 \text{ mol CH}_4$

Why this step? ΔH° is per mole of reaction as written. Need moles to scale.

Step 2: Calculate heat released $q_P = n \times \Delta H° = 1.0 \times (-890.3) = -890.3 \text{ kJ}$

Why negative? Exothermic: system loses 890.3 kJ to surroundings.

Answer: 890.3 kJ released (heats surroundings)

Worked example Example 2: Thermal Decomposition

$\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g) \quad \Delta H° = +178 \text{ kJ/mol}$

Question: A student heats CaCO₃ in an open crucible. Is this spontaneous at 25°C?

Analysis: Step 1: Identify the energy requirement $\Delta H° > 0 \Rightarrow \text{endothermic}$ Why does this matter? System needs continuous heat input from surroundings.

Step 2: Consider spontaneity At 25°C, this reaction is non-spontaneous (ΔG > 0) because:

Large positive ΔH (unfavorable)
Entropy increase from gas formation helps, but not enough at low T

Why this step? Spontaneity depends on ΔG = ΔH - TΔS, not just ΔH.

Step 3: At high temperature (>800°C), TΔS term dominates → ΔG becomes negative → spontaneous

Answer: Not spontaneous at 25°C; requires heating to ~1000 K

Worked example Example 3: Relating ΔH and ΔU

For the reaction: $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g) \quad \Delta H° = -92.4 \text{ kJ/mol}$

Question: Calculate ΔU° at 298 K.

Solution:

Step 1: Use the relationship $\Delta H = \Delta U + \Delta(PV)$

For ideal gases at constant T and P: $\Delta(PV) = \Delta n_{gas} RT$

Why? From ideal gas law: PV = nRT, so Δ(PV) = Δn·RT when T is constant.

Step 2: Calculate Δn_gas $\Delta n_{gas} = n_{\text{products}} - n_{\text{reactants}} = 2 - (1+3) = -2 \text{ mol}$ Why this step? Gas moles decrease, so system does less expansion work.

Step 3: Calculate Δ(PV) $\Delta(PV) = (-2)(8.314)(298) = -4955 \text{ J} = -4.96 \text{ kJ}$

Step 4: Solve for ΔU $\Delta U = \Delta H - \Delta(PV) = -92.4 - (-4.96) = -87.4 \text{ kJ/mol}$

Why is ΔU less negative? Some of the enthalpy decrease comes from reduced gas volume (compression work on system), not just internal energy decrease.

Answer: ΔU° = -87.4 kJ/mol

Common Mistakes

Memory Aids

Active Recall Practice

Recall Feynman Technique: Explain to a 12-year-old

Imagine you have a water balloon. The water inside is like the internal energy U — all the jiggly motion of molecules. Now, the balloon is blown up to a certain size, pushing against the air outside. That "pushing energy" needed to keep the balloon inflated is like the PV part.

Enthalpy is the total: the water's energy PLUS the energy stored in keeping the balloon pushed out against air pressure.

When you do a chemical reaction in an open beaker (like mixing baking soda and vinegar), it happens at normal air pressure — just like our balloon being in the room. The heat you feel (the beaker getting warm or cold) is exactly the enthalpy change ΔH.

Why? Because at constant air pressure, all the heat flow you measure includes both the internal energy change AND the work of any gas bubles expanding. Enthalpy bundles those together!

Exothermic (ΔH negative) = your reaction is like a deflating balloon, releasing warmth. Endothermic (ΔH positive) = your reaction sucks in heat from the room, like inflating a balloon requires you to blow (add energy).

Connections Internal Energy U and First Law — H is defined in terms of U

PV Work and Expansion — Why PV appears in enthalpy definition
Hess's Law — ΔH is a state function, path-independent
Standard Enthalpy of Formation — Building ΔH_rxn from tabulated data
Calorimetry — Experimental measurement of ΔH
Entropy and Gibs Free Energy — ΔH is one piece of spontaneity (ΔG = ΔH - TΔS)
Bond Enthalpies — Estimating ΔH from bond energies

Flashcards

#flashcards/chemistry

What is the definition of enthalpy H? :: H = U + PV, where U is internal energy, P is pressure, V is volume

At constant pressure, what does ΔH equal?

ΔH = q_P (heat exchanged with surroundings at constant pressure)

Why is enthalpy more useful than internal energy for constant-pressure reactions?

At constant P, ΔH directly equals measurable heat flow, automatically accounting for PV work

What is the sign of ΔH for an exothermic reaction?

ΔH < 0 (system releases heat to surroundings)

What is the sign of ΔH for an endothermic reaction?

ΔH > 0 (system absorbs heat from surroundings)

For the reaction N₂ + 3H₂ → 2NH₃, if ΔH = -92 kJ/mol, is ΔU more or less negative?

Less negative (ΔU ≈ -87 kJ/mol) because Δn_gas < 0 means compression work partially offsets internal energy decrease

What is the relationship between ΔH and ΔU for ideal gas reactions?

ΔH = ΔU + Δn_gas·RT, where Δn_gas = (moles of gaseous products) - (moles of gaseous reactants)

If a beaker feels hot after reaction, is ΔH positive or negative?

ΔH is negative (exothermic); system released heat to surroundings, warming the beaker

Under what condition is ΔH NOT equal to heat flow?

When pressure is not constant (e.g., constant volume processes where q_V = ΔU instead)

What thermodynamic quantity does a bomb calorimeter measure?

ΔU (internal energy change at constant volume)

What thermodynamic quantity does a coffee-cup calorimeter measure?

ΔH (enthalpy change at constant pressure)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, enthalpy (H) ek bahut important concept hai thermochemistry mein. Jab bhi tum koi reaction perform karte ho open beaker mein, like mixing acids ya burning fuel, toh woh constant pressure pe hota hai (atmospheric pressure). Enthalpy exactly yeh bata hai kitni heat absorb ya release hui.

Formula simple hai: H = U + PV. Yahan U matlab internal energy (molecules ki total energy), aur PV matlab expansion work ka contribution. Jab reaction hota hai aur gases form hoti hain ya consume hoti hain, toh system ko pressure ke against expand ya contract karna padta hai. Yeh work term automatically enthalpy mein include ho jata hai, isliye constant pressure peΔH directly equals heat flow (qₚ).

Exothermic reaction (ΔH negative) mein beaker garam feel hoga kyunki system ne heat release kari. Endothermic reaction (ΔH positive) mein beaker thanda lagega kyunki system ne surroundings se heat absorb kari. Yeh concept practical lab work mein bahut useful hai — calorimetry experiments mein directly ΔH measure kar sakte ho bina PV work ki tension liye. Chemistry thermodynamics ka yeh foundation piece hai, aur isse samajh ke age Hess's Law, Gibs energy, sab connect ho jata hai!