Question

Consider the second-order reaction:

2HI(g)→H2(g)+I2(g)

Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 4.53×1010 s ([HI]t) for a reaction starting under the condition in the simulation?

Given from simulation:

Rate Law: k[HI]^2

k= 6.4 x 10^-9 l/(mol x s) at 500K

Initial Rate= 1.6 x 10^-7 mol/(l x s)

Answer 1

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)$

Rate Law: $k[HI]^2$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI =$[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.