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.


  • Orange red crystals of K2Cr2Oare 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 (ΔHf) = Standard enthalpy of formation (product – reactant)

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


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.