Question

50.0 g of N2O4(g) is placed in a 2.0 L evacuated flask and the system is allowed to reach equilibrium according to the reaction: N2O4(g) ⇄ 2 NO2(g) Kc = 0.133 After the system reaches equilibrium, 5.00 g of NO2(g) is injected into the vessel, and the system is allowed to equilibrate once again. Calculate the mass of N2O4 in the final equilibrium mixture.

Answer 1

Answer:

Approximately $\rm 43.7\; g$.

Assumption: there's no temperature change.

Explanation:

Refer to a modern periodic table for the relative atomic mass data:

• N: $14.007$;
• O: $15.999$.

$M(\rm N_2O_4) = 14.007 \times 2 + 15.999 \times 4 = \rm 92.010\; g\cdot mol^{-1}$.

$M(\rm NO_2) = 14.007 + 2\times 15.999 = \rm 46.005\; g\cdot mol^{-1}$.

Number of moles of $\rm N_2O_4$ in that sample of $\rm 50.0\;g$:

$\displaystyle n(\mathrm{N_2O_4}) = \frac{m(\mathrm{N_2O_4}) }{M(\mathrm{N_2O_4})} = \rm 0.543419\; mol$.

Concentration of $\mathrm{N_2O_4}$ in that $\rm 2.0\; L$ container:

$\displaystyle c = \frac{n}{V} = \rm 0.271710\; mol\cdot L^{-1}$.

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 $\rm N_2O_4\; (g)$ be $-x\; \rm mol\cdot L^{-1}$.

$\begin{array}{c|ccc}\mathbf{R} & \mathrm{N_2O_4}\; (g) & \rightleftharpoons & 2\;\mathrm{NO_2}\;(g)\\\mathbf{I} & 0.271710 & & \\ \mathbf{C} & -x & & +2\;x \\\mathbf{E} & 0.271710 -x & & 2x\end{array}$.

The equilibrium concentrations shall satisfy the equilibrium law for this reaction under this particular temperature.

$\displaystyle \frac{[\rm NO_2]^{2}}{[\rm N_2O_4]} = \rm K_{c} = 0.133$.

$\displaystyle \frac{(2x)^{2}}{0.271710 - x} = 0.133$.

This equation can be simplified to a quadratic equation. Solve this equation. Note that there might be more than one possible values for $x$. $x$ 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 $x$-value.

$x \approx 0.0798672$.

$\rm [N_2O_4] \approx 0.271710 - 0.0798672 = \rm 0.191843 \;mol\cdot L^{-1}$.

$\rm [NO_2] \approx 2 \times 0.0798672 = 0.159734\; mol\cdot L^{-1}$.

What will be the concentration of that additional $\rm 5.00\; g$ of $\rm[NO_2]$ if it was added to an evacuated $\rm 2.0\; L$ flask?

$\displaystyle n(\mathrm{NO_2}) = \frac{m(\mathrm{NO_2}) }{M(\mathrm{NO_2})} = \rm 1.08684\; mol$.

$\displaystyle c = \frac{n}{V} = \frac{1.08684}{2.0} = \rm 0.543419\;mol\cdot L^{-1}$.

The new concentration of $\mathrm{NO_2}\; (g)$ will become

$0.543419 + 0.159734 = \rm 0.703154\; mol\cdot L^{-1}$.

Construct another RICE table. Let the change in the concentration of $\mathrm{N_2O_4}\; (g)$ be $-x\;\rm mol\cdot L^{-1}$.

$\begin{array}{c|ccc}\mathbf{R} & \mathrm{N_2O_4}\; (g) & \rightleftharpoons & 2\;\mathrm{NO_2}\;(g)\\\mathbf{I} & 0.191843 & & 0.703154 \\ \mathbf{C} & -x & & +2\;x \\\mathbf{E} & 0.191843 -x & & 0.703154 + 2x\end{array}$.

Once again, the equilibrium conditions shall satisfy this particular equilibrium law.

$\displaystyle \frac{[\rm NO_2]^{2}}{[\rm N_2O_4]} = \rm K_{c} = 0.133$

$\displaystyle \frac{(0.703154 + 2x)^{2}}{0.191843 -x} = 0.133$.

Simplify and solve this equation for $x$. Make sure that the $x$-value ensures that all concentrations are positive.

$x \approx \rm -0.232758$.

Note that $x \neq -0.503646$ for if that would lead to a negative value for the concentration of $\mathrm{NO}_2$.

Hence the equilibrium concentration of $\rm N_2O_4$ will be:

$[{\rm N_2O_4}] = 0.703154 + 2(-0.232758) = \rm 0.237638\; mol\cdot L^{-1}$.

$n = c\cdot V = 0.237638\times 2 = \rm 0.475276\; mol\cdot L^{-1}$.

$m = n \cdot M = 0.475276 \times 92.010 \approx \rm 43.7\; g$