Respuesta :
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
[tex]Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}[/tex]
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.

The initial concentration of the HI at the time 0 sec, has been 0.05 mol/L.
The balanced chemical equation for the reaction has been:
[tex]\rm H_2\;+\;I_2\;\rightarrow\;2\;HI[/tex]
The given initial concentration of hydrogen has been 0.030M, and the given concentration of iodine has been 0.015 M.
The reaction quotient for the following reaction has been:
Reaction quotient = [tex]\rm \dfrac{[HI]^2}{[H_2]\;[I_2]}[/tex]
The concentration of HI for the reaction quotient 5.56 has been:
5.56 = [tex]\rm \dfrac{[HI]^2}{[0.030]\;[0.015]}[/tex]
[tex]\rm [HI]^2[/tex] = 0.0025 M
HI = 0.05 mol/L or 0.05 M.
The concentration of the HI at the initial concentration has been 0.05 mol/L.
The graph attached has been plotted as the concentration as the function of time.
The initial concentration of hydrogen, iodine, and hydrogen iodide have been plotted as the function of time.
For more information about the reaction quotient, refer to the link:
https://brainly.com/question/8205004
