Is conductivity a chemical or physical property
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Step 1 : Write balanced chemical equation
The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.
Step 2 : Set up ICE table
We will set up an ICE table for the above reaction
Following points are considered while drawing ICE table
- The initial concentration of product is assumed as 0
- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products
- The coefficients in balanced equation are considered while writing C values
Check attached file for ICE table
Step 3 : Set up equilibrium constant equation
The equation for equilibrium constant can be written as
Step 4 : Solving for x
Keq at 298.15 K is given as 81.9
Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table
81.9 =
81.9 =
Solving the above equation using quadratic formula we get
x = 0.488 or x = 0.254
The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.
Therefore the change in concentration of the gases during the reaction is 0.254 M
Hence, x = 0.254 M
From the ICE table, we know that the equilibrium concentration of ICl is 2x
The concentration of ICl when the reaction reaches equilibrium is 0.508 M
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>How to find the empirical formula?</h3>Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus = 0.0293 mol
Moles bromine 6.99 g bromine=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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