Enthalpy Calculator

Find heat, heat capacity and reaction enthalpy with AI-powered step-by-step solutions
q for 150.0 g of water heated from 22.0 C to 78.0 C
Enthalpy of combustion of CH4 from standard formation enthalpies
Molar enthalpy from 2717 J released by 0.0500 mol
Specific heat capacity from q, mass and temperature change

Enthalpy and Heat

Enthalpy HH is the heat content of a system at constant pressure. Only changes in it are measurable, and at constant pressure the enthalpy change equals the heat exchanged:

ΔH=qp\Delta H = q_p

The sign convention is the system's point of view: ΔH<0\Delta H < 0 is exothermic (heat leaves the system), ΔH>0\Delta H > 0 is endothermic.

Heating without a reaction. When a substance simply warms or cools,

q=mcΔTq = m c \Delta T

  • mm — mass in g, ccspecific heat capacity in J g⁻¹ °C⁻¹ (4.184 for liquid water), ΔT=TfinalTinitial\Delta T = T_{\text{final}} - T_{\text{initial}} in °C or K — the size of a degree is the same on both scales, so no conversion is needed for a difference.

Reaction enthalpy from formation data. With tabulated standard enthalpies of formation ΔHf\Delta H_f^{\circ},

ΔHrxn=νΔHf(products)νΔHf(reactants)\Delta H^{\circ}_{\text{rxn}} = \sum \nu\,\Delta H_f^{\circ}(\text{products}) - \sum \nu\,\Delta H_f^{\circ}(\text{reactants})

where ν\nu are the balanced coefficients. Standard state means 1 bar with each substance in its normal form at the stated temperature, usually 298.15 K, and ΔHf\Delta H_f^{\circ} of an element in its standard state is exactly zero. Physical states matter: liquid and gaseous water differ by about 44 kJ/mol.

Three Enthalpy Calculations

1. Heat from a temperature change

Multiply mass, specific heat and ΔT\Delta T. Keep the sign of ΔT\Delta T: a negative value means heat was released. If a heat capacity CC for a whole object is given instead of a specific heat, use q=CΔTq = C\Delta T with no mass.

2. Molar enthalpy from calorimetry data

A calorimetry experiment reports how much a known mass of solution changed temperature. Compute the heat absorbed by that solution, flip the sign to get the heat released by the reaction, then divide by the moles of the limiting reactant:

ΔH=qsolutionn\Delta H = \frac{-q_{\text{solution}}}{n}

3. Reaction enthalpy from tables or Hess's law

Sum the formation enthalpies of the products, subtract those of the reactants, and weight each by its coefficient. Hess's law is the same principle stated generally: because enthalpy is a state function, the total change is the same whatever route is taken, so known reactions can be reversed (flip the sign) and scaled (multiply the value) and then added.

Units and significant figures

Answers usually convert from J to kJ, and reaction enthalpies are per mole of reaction as written. The measured mass or temperature change normally sets the significant figures; 4.184 is quoted to four.

Common Mistakes to Avoid

  • Dropping the sign. Exothermic reactions have negative ΔH\Delta H. Reporting +890+890 kJ/mol for a combustion reverses the physics.
  • Reversing ΔT\Delta T. It is final minus initial. Getting it backwards flips the sign of qq.
  • Mixing J and kJ. q=mcΔTq = mc\Delta T with cc in J g⁻¹ °C⁻¹ gives joules; formation enthalpies are tabulated in kJ/mol.
  • Forgetting to divide by moles. qq from a calorimeter is for the whole sample; ΔH\Delta H in kJ/mol needs division by the moles of limiting reactant.
  • Ignoring physical states. ΔHf\Delta H_f^{\circ} for H2O(l)\mathrm{H_2O}(l) and H2O(g)\mathrm{H_2O}(g) differ, and the products' states change the answer.
  • Forgetting to scale a Hess's law step. Doubling a reaction doubles its ΔH\Delta H; reversing it changes the sign.
  • Assuming ΔHf\Delta H_f^{\circ} of any element is zero. Only the standard form counts: O2(g)\mathrm{O_2}(g) is zero, ozone is not.

Examples

Step 1: ΔT=78.022.0=56.0\Delta T = 78.0 - 22.0 = 56.0 °C
Step 2: q=mcΔT=(150.0)(4.184)(56.0)q = mc\Delta T = (150.0)(4.184)(56.0)
Step 3: (150.0)(4.184)=627.6 JC1(150.0)(4.184) = 627.6\ \mathrm{J\,^{\circ}C^{-1}}
Step 4: q=627.6×56.0=35146q = 627.6 \times 56.0 = 35146 J
Step 5: Convert and round to the 3 significant figures of ΔT\Delta T: q=35.1q = 35.1 kJ, positive because heat was absorbed
Answer: q=+35.1q = +35.1 kJ

Step 1: Products: (393.5)+2(285.8)=393.5571.6=965.1(-393.5) + 2(-285.8) = -393.5 - 571.6 = -965.1 kJ
Step 2: Reactants: (74.6)+2(0)=74.6(-74.6) + 2(0) = -74.6 kJ, since O2(g)\mathrm{O_2}(g) is an element in its standard state
Step 3: ΔH=965.1(74.6)=965.1+74.6\Delta H^{\circ} = -965.1 - (-74.6) = -965.1 + 74.6
Step 4: ΔH=890.5\Delta H^{\circ} = -890.5 kJ per mole of CH4\mathrm{CH_4}
Step 5: The negative sign confirms an exothermic reaction
Answer: ΔH=890.5\Delta H^{\circ} = -890.5 kJ/mol

Step 1: Heat gained by the solution: q=(100.0)(4.18)(6.50)=2717q = (100.0)(4.18)(6.50) = 2717 J
Step 2: The solution gained that heat, so the reaction released it: qrxn=2717q_{\text{rxn}} = -2717 J
Step 3: Per mole: ΔH=2717 J0.0500 mol=54340 J/mol\Delta H = \dfrac{-2717\ \mathrm{J}}{0.0500\ \mathrm{mol}} = -54340\ \mathrm{J/mol}
Step 4: Convert to kJ and round to 3 significant figures: 54.3-54.3 kJ/mol
Answer: ΔH=54.3\Delta H = -54.3 kJ/mol

Frequently Asked Questions

At constant pressure ΔH equals the heat exchanged, q. For simple heating, q = mcΔT. For a reaction with tabulated data, ΔH°rxn = sum of coefficient x ΔHf° for the products minus the same sum for the reactants.

Specific heat capacity c is per gram, in J g⁻¹ °C⁻¹, so q = mcΔT. Heat capacity C is for a whole object, in J/°C, so q = CΔT with no mass term. Molar heat capacity is per mole instead of per gram.

Because the sign is written from the system's point of view. An exothermic reaction sends heat out to the surroundings, so the system's enthalpy falls and ΔH is negative. Methane combustion is -890.5 kJ per mole of methane.

Enthalpy is a state function, so ΔH depends only on the initial and final states, not the route. You can therefore add known reactions to build the target one — reversing a step flips the sign of its ΔH, and scaling a step multiplies its ΔH by the same factor.

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