The data below refer to the following reaction: 2NO(g) + I2(g) 2NOI(g) Concentration (M) [NO] [I2] [NOI] Initial 2.0 4.0 1.0 Equilibrium 1.0 — — Find the concentration of I2 when the system reaches equilibrium.

Answer:

3.5 mol·L⁻¹  

Explanation:

1. Set up an ICE table.

[tex]\begin{array}{cccccc}\text{2NO} & + & \text{I}_{2} &\, \rightleftharpoons \, & \text{2NOI} & & \\ 2.0 & & 4.0 & & 1.0 & & \\ -2x & & -x & & +2x & & \\ 2.0-2x & & 4.0-x & & 1.0+2x & & \\\end{array}[/tex]

2. Solve for x

The equilibrium concentration of NO is 1.0 mol·L⁻¹, so

       1.0 = 2.0 - 2x

2x + 1.0 = 2.0

        2x =  1.0

          x = 0.5

3. Calculate the equilibrium concentration of I₂

[I₂] = 4.0 - x = 4.0 - 0.5 = 3.5 mol·L⁻¹

The concentration of I₂ at equilibrium is calculated to be 3.5 M by using the initial concentration of NO to determine the stoichiometric change in I₂ concentration based on the reaction 2NO(g) + I₂(g) → 2NOI(g).

To find the concentration of I₂ at equilibrium for the reaction 2NO(g) + I₂(g) → 2NOI(g), we use the initial and equilibrium concentrations of NO to determine the change in concentration of I₂. Given the stoichiometry of the reaction, for every 1 mole decrease in NO, there is a 0.5 mole decrease in I₂. The initial concentration of NO is 2.0 M, and at equilibrium, it is 1.0 M, which means there has been a 1.0 M decrease (2.0 M - 1.0 M). The I₂ concentration at equilibrium can be found by subtracting half of this change from the initial I₂ concentration. Since initially the concentration of I₂ is 4.0 M, the equilibrium concentration is calculated as 4.0 M - (1.0 M / 2) = 3.5 M.

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