15 Facts on HBr + K2Cr2O7: What, How To Balance & FAQs

The reaction between HBr and K₂Cr₂O₇  is an oxidation-reduction reaction. Let us see through this article how HBr and K₂Cr₂O₇ react

HBr exists in liquid form whereas K₂Cr₂O₇ exists in the form of orange red crystals which when dissolved in water is used as a reagent in analytical chemistry for estimation of ions in volumetric analysis. It is also used as an oxidizing agent in organic chemistry.

This article describes various facts about reaction between HBr + K₂Cr₂O₇ like products formed, type of reaction, balancing, titration, net ionic equation, enthalpy of reaction, intermolecular forces involved, etc.

What is the product of HBr + K₂Cr₂O₇?

When HBr reacts with K₂Cr₂O₇ it gives, potassium bromide ( KBr), Chromium(III) bromide (CrBr₃) , bromine (Br₂) molecule and water ( H₂O)respectively. The complete balanced reaction is written as:

K₂Cr₂O₇ + 14 HBr = 2KBr + 2CrBr₃ + 3Br₂ + 7H₂O

What type of reaction is HBr + K₂Cr₂O₇?

HBr + K₂Cr₂O₇ is a dissociation reaction where products dissociate into ions in solution.

How to balance HBr + K₂Cr₂O₇?

The HBr + K₂Cr₂O₇ equation is balanced by following the below given steps:

• The unbalanced equation between HBr + K₂Cr₂O₇ is written as follows-
  • K₂Cr₂O₇ + HBr = KBr + CrBr₃ + Br₂ + H₂O
• For balancing K and Cr, we need to multiply KBr and CrBr₃ with 2 on the right hand side so we get-
  • K₂Cr₂O₇ + HBr = 2 KBr + 2CrBr₃ + Br₂ + H₂O
• Balance the O atoms by multiplying water with 7 on the right hand side-
  • K₂Cr₂O₇ + HBr = 2 KBr +2 CrBr₃ + Br₂ +7 H₂O
• Now we have 14 H atoms on the product side, which can be equated on the reactant side by multiplying HBr with 14-
  • K₂Cr₂O₇ +14 HBr = 2 KBr + 2CrBr₃ + Br₂ +7 H₂O
• To balance the chemical equation multiply Br atoms on the right hand side with 3-
  • K₂Cr₂O₇ +14 HBr = 2 KBr + 2CrBr₃ + 3Br₂ +7 H₂O

HBr + K₂Cr₂O₇ Titration

HBr + K₂Cr₂O₇ titration involes following steps-

Appartus used

• Graduated burette
• Conical flask
• Volumetric flask
• Burette Stand
• Beakers
• Graduated pipette

Titre and Titrant

• Titrant is a substance having known concentration. In this HBr + K₂Cr₂O₇ titration, K₂Cr₂O₇ is used as a titrant.
• Titre or analyte is a substance whose concentration is to be determined. In this reaction, HBr is used as a titre.

Procedure

• Orange red crystals of K₂Cr₂O₇  are weighed and a known standard solution of K₂Cr₂O₇ is prepared in the volumetric flask.
• Unknown Concentration of solution of HBr is filled in the burette and clamped to the burette stand.
• A known volume of K₂Cr₂O₇ is taken in the conical flask which is then titrated with drops of HBr from the burette.
• The change in colour of K₂Cr₂O₇ shows the end point of titration.

HBr + K₂Cr₂O₇ Net Ionic equation

The net ionic equation of the reaction is –

K₂Cr₂O₇(s) +14 HBr(aq)  = 2 K⁺ (aq) + 2Cr³⁺(aq) + 8Br^– (aq) + 3Br₂(g) + 7 H₂O(l)

HBr + K₂Cr₂O₇ reaction gives reddish brown fumes of bromine molecule while in the aqueous solution, potassium, chromium(III) and bromide ions are present.

HBr + K₂Cr₂O₇ Conjugate pairs

In the reaction HBr + K₂Cr₂O₇, K₂Cr₂O₇ does not have any conjugate pairs as they are formed by loss or gain of a proton.

• Conjugate base of HBr = Br^–
• Conjugate acid of H2O = OH^–  

HBr + K₂Cr₂O₇ intermolecular forces

• HBr is a polar molecule having dipole – dipole forces.
• HBr shows interionic attractions in an aqueous solution.
• Potassium dichromate dissociates into chromate and potassium ions.

HBr + K₂Cr₂O₇ Reaction Enthalpy

The enthalpy of reaction for HBr + K₂Cr₂O₇ is –708.6 kJ/ mol. This can be calculated using the enthalpy of formation of different reactants and products which are given as:

• Enthaply of formation of K₂Cr₂O₇ = -2035 kJ/mole
• Enthaply of formation of HBr = – 36.2 kJ/ mol
• Enthalpy of formation of Br₂ = 111.8 kJ/mol
• Enthalpy of formation of CrBr₃ = – 400.4 kJ/mol
• Enthalpy of formation of H₂O = – 285.8 kJ/mol
• Enthalpy of formation of KBr = – 392.2 kJ/mol

Reaction enthalpy (ΔH_f) = Standard enthalpy of formation (product – reactant)

Thus, ΔH_f = [2*(-392.2) + 3*(111.8) + 2*(-400.4) + 7*(-285.8)] – [(-2035) + 14*(-36.2)]
ΔH_f = [ -3250.4] – [+2541.8]
ΔH_f = -708.6 kJ/mol.

Is HBr + K₂Cr₂O₇ a Buffer solution?

HBr + K₂Cr₂O₇ do not form a buffer solution as HBr is a strong acid and dissociates completely in solution at all concentrations while buffer solutions as formed by weak acids or bases.

Is HBr + K₂Cr₂O₇ a Complete reaction?

The reaction between HBr + K₂Cr₂O₇ is a complete reaction and proceeds in one direction only as the products are completely formed.

Is HBr + K₂Cr₂O₇ an Exothermic reaction or Endothermic reaction?

Is HBr + K₂Cr₂O₇  is an exothermic reaction having negative enthalpy of reaction which is given by -708.6 kJ/mol

Is HBr + K₂Cr₂O₇ a Redox reaction?

The reaction between HBr and  K₂Cr₂O₇  is redox reaction involving reduction of K₂Cr₂O₇ where oxidation state of chromium changes from +6 to +3 and oxidation of HBr, where oxidation state of bromine changes from -1 to 0. In this redox reaction, HBr acts as a reducing agent while K₂Cr₂O₇ behaves as an oxidizing agent.

Is HBr + K₂Cr₂O₇ a Precipitation reaction?

HBr + K₂Cr₂O₇  is not a precipitation reaction as no precipitates ( solid phase compounds) are formed at the end of the reaction.

Is HBr + K₂Cr₂O₇ a Reversible reaction?

 HBr + K₂Cr₂O₇  is not a reversible reaction as the reaction proceeds towards the right hand side only leading to the formation of products.

Is HBr + K₂Cr₂O₇ a Displacement reaction?

HBr and K₂Cr₂O₇ is not a displacement reaction as no element is replaced by another in the reaction.

Conclusion

The reaction between HBr + K₂Cr₂O₇  forms blackish chromium(III) bromide along with reddish brown irritating fumes of bromine gas. This reaction presents a significant example of redox reaction showing change in colour.