What type of reaction is hi H2 I2?

What type of reaction is hi H2 I2?

This is a second order reaction.

Is I to I2 oxidation or reduction?

The iodine atoms go from -1 to 0. This is an increase in oxidation number, so the iodine is oxidized, and we can write the oxidation half of the overall equation: I- → I2.

Is 2hi → H2 I2 a redox reaction?

The reaction is a reduction-oxidation reaction. The reaction involves the decomposition of hydroiodic acid into hydrogen gas and iodine gas.

What is HI Chem?

hydrogen iodide (HI), known as hydroiodic acid, is a strong acid that is used to prepare iodides by reaction with metals or their oxides, hydroxides, and carbonates.

What is 2HI in chemistry?

Hydroiodic acid (aqueous solution) Iodine hydride. Identifiers. CAS Number. 10034-85-2.

What type of reaction is HI?

Chemical Reversibility One example of a reversible reaction is the reaction of hydrogen gas and iodine vapor to form hydrogen iodide.

What is the chemical reaction for HI?

HI is a colorless gas that reacts with oxygen to give water and iodine. With moist air, HI gives a mist (or fumes) of hydroiodic acid. It is exceptionally soluble in water, giving hydroiodic acid.

What is the name of 2I I2 reaction?

Conversion of 2I to I2 is (oxidation /reduction).

What is the oxidation number of I in I2?

The oxidation number of I in the iodine molecule is 0 .

Is H2 I2 2HI exothermic or endothermic?

H2(g)+I2(g)⇌2HI(g)ΔH=−9.4kJ(exothermic) Because this reaction is exothermic, we can write it with heat as a product. Your Response.

Is HI an element or compound?

iodine compounds hydrogen iodide (HI), known as hydroiodic acid, is a strong acid that is used to prepare iodides by reaction with metals or their oxides, hydroxides, and carbonates.

What is the formula for Hi from H2 and I2?

Formation of HI from H2 and I2 The formation of HI from H2 and I2 is an example of gaseous homogeneous equilibrium reaction. It can be represented as H2 (g) + I2 (g)– > < — 2HI(g) ∆H=-10.4 kJ

How do you calculate KC from H2 and I2?

According to the law of mass action, Kc = [HI]2 / [H2] [I2] Substituting the values of equilibrium concentrations in the above equation, we get. Kc = 1/ (1-x)2. If the initial concentration of H2 and I2 are equal to a and b moles dm-3 respectively, then it can be shown that.

How to calculate the concentrations of H2 I2 and Hi at equilibrium?

The concentrations of H2, I2 and HI remaining at equilibrium can be calculated as follows : Substituting the values of equilibrium concentrations in the above equation, we get If the initial concentration of H2 and I2 are equal to a and b moles dm-3 respectively, then it can be shown that

What is the half-equation for oxidation of iodide ions to form iodine?

Construct the half-equation for the oxidation of iodide ions to form iodine”. I- –>I2. So it went from -1 to 0. Hence it lost an electron. I- –> I2 + e-. Balance the equation to get: 2I- –>I2 +2e-. 0. reply.