How To Calculate Change In H Of A Reaction

Rut of Reaction

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The Heat of Reaction (also known and Enthalpy of Reaction) is the change in the enthalpy of a chemic reaction that occurs at a abiding pressure. It is a thermodynamic unit of measurement useful for calculating the corporeality of free energy per mole either released or produced in a reaction. Since enthalpy is derived from pressure, volume, and internal free energy, all of which are state functions, enthalpy is also a state function.

Introduction

\(ΔH\), or the change in enthalpy arose every bit a unit of measurement meant to summate the change in energy of a organization when information technology became too difficult to discover the ΔU, or modify in the internal energy of a organization, by simultaneously measure the corporeality of heat and work exchanged. Given a constant pressure, the change in enthalpy can be measured as

\[ΔH=q\]

Encounter section on enthalpy for a more than detailed explanation.

The notation ΔHº or ΔHº_rxn then arises to explain the precise temperature and pressure of the heat of reaction ΔH. The standard enthalpy of reaction is symbolized by ΔHº or ΔHº_rxn and can accept on both positive and negative values. The units for ΔHº are kiloJoules per mole, or kj/mol.

ΔH and ΔH^º _rxn

Δ = represents the alter in the enthalpy; (ΔH_products -ΔH_reactants)
- a positive value indicates the products have greater enthalpy, or that information technology is an endothermic reaction (heat is required)
- a negative value indicates the reactants accept greater enthalpy, or that it is an exothermic reaction (heat is produced)
º = signifies that the reaction is a standard enthalpy alter, and occurs at a preset pressure/temperature
_rxn = denotes that this alter is the enthalpy of reaction

The Standard State: The standard country of a solid or liquid is the pure substance at a pressure of 1 bar ( 10⁵ Pa) and at a relevant temperature.

The ΔHº_rxn is the standard rut of reaction or standard enthalpy of a reaction, and like ΔH too measures the enthalpy of a reaction. However, ΔHº_rxn takes place under "standard" conditions, meaning that the reaction takes place at 25º C and ane atm. The benefit of a measuring ΔH under standard conditions lies in the ability to relate one value of ΔHº to another, since they occur nether the same conditions.

How to Calculate ΔH Experimentally

Enthalpy can be measured experimentally through the employ of a calorimeter. A calorimeter is an isolated organization which has a constant pressure, so ΔH=q=cp_sp x m x (ΔT)

How to calculate ΔH Numerically

To calculate the standard enthalpy of reaction the standard enthalpy of formation must be utilized. Another, more detailed, form of the standard enthalpy of reaction includes the use of the standard enthalpy of germination ΔH^º _f:

\[ ΔH^\ominus = \sum \Delta v_p \Delta H^\ominus_f\;(products) - \sum \Delta v_r \Delta H^\ominus_f\; (reactants)\]

with

v_p= stoichiometric coefficient of the product from the balanced reaction
v_r= stoichiometric coefficient of the reactants from the balanced reaction
ΔH^º _f= standard enthalpy of germination for the reactants or the products

Since enthalpy is a state function, the heat of reaction depends merely on the terminal and initial states, not on the path that the reaction takes. For example, the reaction \( A \rightarrow B\) goes through intermediate steps (i.due east. \(C \rightarrow D\)), but A and B remain intact.

Therefore, one can measure the enthalpy of reaction as the sum of the ΔH of the 3 reactions by applying Hess' Constabulary.

Additional Notes

Since the ΔHº represents the full energy exchange in the reaction this value tin can be either positive or negative.

A positive ΔHº value represents an addition of energy from the reaction (and from the surroundings), resulting in an endothermic reaction.
A negative value for ΔHº represents a removal of energy from the reaction (and into the surroundings) and then the reaction is exothermic.

Case \(\PageIndex{1}\): the combustion of acetylene

Summate the enthalpy change for the combustion of acetylene (\(\ce{C2H2}\))

Solution

1) The first step is to make sure that the equation is balanced and correct. Recollect, the combustion of a hydrocarbon requires oxygen and results in the product of carbon dioxide and h2o.

\[\ce{2C2H2(chiliad) + 5O2(thousand) -> 4CO2(k) + 2H2O(yard)}\]

2) Side by side, locate a tabular array of Standard Enthalpies of Formation to look upward the values for the components of the reaction (Table vii.2, Petrucci Text)

iii) Get-go find the enthalpies of the products:

ΔHº_f CO_ii = -393.5 kJ/mole

Multiply this value by the stoichiometric coefficient, which in this case is equal to 4 mole.

v_pΔH^º _f CO₂ = 4 mol (-393.5 kJ/mole)

= -1574 kJ

ΔH^º _f H₂O = -241.viii kJ/mole

The stoichiometric coefficient of this compound is equal to 2 mole. Then,

v_pΔH^º _f H_iiO = 2 mol ( -241.8 kJ/mole)

= -483.6 kJ

Now add these two values in club to get the sum of the products

Sum of products (Σ v_pΔHº_f(products)) = (-1574 kJ) + (-483.half dozen kJ) = -2057.6 kJ

Now, find the enthalpies of the reactants:

ΔHº_f C₂H_two = +227 kJ/mole

Multiply this value by the stoichiometric coefficient, which in this case is equal to ii mole.

v_pΔHº_f C₂H_ii = 2 mol (+227 kJ/mole)

= +454 kJ

ΔHº_f O_two = 0.00 kJ/mole

The stoichiometric coefficient of this compound is equal to v mole. And so,

v_pΔHº_f O₂ = 5 mol ( 0.00 kJ/mole)

= 0.00 kJ

Add these two values in social club to get the sum of the reactants

Sum of reactants (Δ five_rΔHº_f(reactants)) = (+454 kJ) + (0.00 kJ) = +454 kJ

The sum of the reactants and products tin at present be inserted into the formula:

ΔHº = Δ 5_pΔHº_f(products) - ? v_rΔHº_f(reactants)

= -2057.six kJ - +454 kJ

= -2511.6 kJ

Exercise Problems

Summate ΔH if a piece of metallic with a specific heat of .98 kJ·kg−ane·M−i and a mass of 2 kg is heated from 22^oC to 28^oC.
If a calorimeter'due south ΔH is +2001 Joules, how much heat did the substance inside the loving cup lose?
Summate the ΔH of the following reaction: CO_{2 (g)}+ H_twoO _{(g) -->}H_iiCO_three _(grand)if the standard values of ΔH_f are equally follows: CO_two _(g): -393.509 KJ /mol, H_twoO_{(one thousand)}: -241.83 KJ/mol, and H₂CO_iii _(g) : -275.ii KJ/mol.
Calculate ΔH if a piece of aluminum with a specific heat of .9 kJ·kg−1·K−i and a mass of 1.6 kg is heated from 286^oK to 299^oK.
If the calculated value of ΔH is positive, does that correspond to an endothermic reaction or an exothermic reaction?

Solutions

ΔH=q=cp_sp x chiliad x (ΔT) = (.98) ten (2) 10 (+6^o) = 11.76 kJ
Since the heat gained past the calorimeter is equal to the oestrus lost by the arrangement, then the substance inside must accept lost the negative of +2001 J, which is -2001 J.
ΔH^º = ∑Δv_pΔH^º _f(products) - ∑Δ v_rΔH^º _f(reactants) so this means that y'all add up the sum of the ΔH'due south of the products and subtract away the ΔH of the products: (-275.2kJ) - (-393.509kJ + -241.83kJ) = (-275.ii) - (-635.339) = +360.139 kJ.
ΔH=q=cp_sp x k x (ΔT) = (.9) x (ane.half dozen) ten (13) = xviii.72 kJ
Endothermic, since a positive value indicates that the arrangement GAINED heat.

References

Petrucci, et al. General Chemical science: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey 2007.
Zumdahl, Steven Southward., and Susan A. Zumdahl. Chemistry. Boston: Houghton Mifflin Company, 2007.

Contributors and Attributions

Rachel Martin (UCD), Eleanor Yu (UCD)