Respuesta :
Answer:
Approximately [tex]\rm 43.7\; g[/tex].
Assumption: there's no temperature change.
Explanation:
Refer to a modern periodic table for the relative atomic mass data:
- N: [tex]14.007[/tex];
- O: [tex]15.999[/tex].
[tex]M(\rm N_2O_4) = 14.007 \times 2 + 15.999 \times 4 = \rm 92.010\; g\cdot mol^{-1}[/tex].
[tex]M(\rm NO_2) = 14.007 + 2\times 15.999 = \rm 46.005\; g\cdot mol^{-1}[/tex].
Number of moles of [tex]\rm N_2O_4[/tex] in that sample of [tex]\rm 50.0\;g[/tex]:
[tex]\displaystyle n(\mathrm{N_2O_4}) = \frac{m(\mathrm{N_2O_4}) }{M(\mathrm{N_2O_4})} = \rm 0.543419\; mol[/tex].
Concentration of [tex]\mathrm{N_2O_4}[/tex] in that [tex]\rm 2.0\; L[/tex] container:
[tex]\displaystyle c = \frac{n}{V} = \rm 0.271710\; mol\cdot L^{-1}[/tex].
Construct a RICE table for this equilibrium. Note that all values in this table shall stand for concentrations. Let the change in the concentration of [tex]\rm N_2O_4\; (g)[/tex] be [tex]-x\; \rm mol\cdot L^{-1}[/tex].
[tex]\begin{array}{c|ccc}\mathbf{R} & \mathrm{N_2O_4}\; (g) & \rightleftharpoons & 2\;\mathrm{NO_2}\;(g)\\\mathbf{I} & 0.271710 & & \\ \mathbf{C} & -x & & +2\;x \\\mathbf{E} & 0.271710 -x & & 2x\end{array}[/tex].
The equilibrium concentrations shall satisfy the equilibrium law for this reaction under this particular temperature.
[tex]\displaystyle \frac{[\rm NO_2]^{2}}{[\rm N_2O_4]} = \rm K_{c} = 0.133[/tex].
[tex]\displaystyle \frac{(2x)^{2}}{0.271710 - x} = 0.133[/tex].
This equation can be simplified to a quadratic equation. Solve this equation. Note that there might be more than one possible values for [tex]x[/tex]. [tex]x[/tex] itself might not necessarily be positive. However, keep in mind that all concentrations in an equilibrium should be positive. Apply that property to check the [tex]x[/tex]-value.
[tex]x \approx 0.0798672[/tex].
[tex]\rm [N_2O_4] \approx 0.271710 - 0.0798672 = \rm 0.191843 \;mol\cdot L^{-1}[/tex].
[tex]\rm [NO_2] \approx 2 \times 0.0798672 = 0.159734\; mol\cdot L^{-1}[/tex].
What will be the concentration of that additional [tex]\rm 5.00\; g[/tex] of [tex]\rm[NO_2][/tex] if it was added to an evacuated [tex]\rm 2.0\; L[/tex] flask?
[tex]\displaystyle n(\mathrm{NO_2}) = \frac{m(\mathrm{NO_2}) }{M(\mathrm{NO_2})} = \rm 1.08684\; mol[/tex].
[tex]\displaystyle c = \frac{n}{V} = \frac{1.08684}{2.0} = \rm 0.543419\;mol\cdot L^{-1}[/tex].
The new concentration of [tex]\mathrm{NO_2}\; (g)[/tex] will become
[tex]0.543419 + 0.159734 = \rm 0.703154\; mol\cdot L^{-1}[/tex].
Construct another RICE table. Let the change in the concentration of [tex]\mathrm{N_2O_4}\; (g)[/tex] be [tex]-x\;\rm mol\cdot L^{-1}[/tex].
[tex]\begin{array}{c|ccc}\mathbf{R} & \mathrm{N_2O_4}\; (g) & \rightleftharpoons & 2\;\mathrm{NO_2}\;(g)\\\mathbf{I} & 0.191843 & & 0.703154 \\ \mathbf{C} & -x & & +2\;x \\\mathbf{E} & 0.191843 -x & & 0.703154 + 2x\end{array}[/tex].
Once again, the equilibrium conditions shall satisfy this particular equilibrium law.
[tex]\displaystyle \frac{[\rm NO_2]^{2}}{[\rm N_2O_4]} = \rm K_{c} = 0.133[/tex]
[tex]\displaystyle \frac{(0.703154 + 2x)^{2}}{0.191843 -x} = 0.133[/tex].
Simplify and solve this equation for [tex]x[/tex]. Make sure that the [tex]x[/tex]-value ensures that all concentrations are positive.
[tex]x \approx \rm -0.232758[/tex].
Note that [tex]x \neq -0.503646[/tex] for if that would lead to a negative value for the concentration of [tex]\mathrm{NO}_2[/tex].
Hence the equilibrium concentration of [tex]\rm N_2O_4[/tex] will be:
[tex][{\rm N_2O_4}] = 0.703154 + 2(-0.232758) = \rm 0.237638\; mol\cdot L^{-1}[/tex].
[tex]n = c\cdot V = 0.237638\times 2 = \rm 0.475276\; mol\cdot L^{-1}[/tex].
[tex]m = n \cdot M = 0.475276 \times 92.010 \approx \rm 43.7\; g[/tex]
