[tex][\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}[/tex] at equilibrium.
Explanation
Concentration for each of the species:
- [tex][\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}[/tex];
- [tex][\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}[/tex];
- [tex][\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}[/tex].
There was no Y to start with; its concentration could only have increased. Let the change in [tex][\text{Y}][/tex] be [tex]+x \; \text{mol}\cdot \text{dm}^{-3}[/tex].
Make a [tex]\textbf{RICE}[/tex] table.
Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the change in [tex][\text{X}][/tex] will be [tex]+2\;x \; \text{mol}\cdot \text{dm}^{-3}[/tex] and the change in [tex][\text{Z}][/tex] will be [tex]-2\;x \; \text{mol}\cdot \text{dm}^{-3}[/tex].
[tex]\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}[/tex].
Add the value in the C row to the I row:
[tex]\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 + 2\;x & &x&&3-2\;x\end{array}[/tex].
What's the equation of [tex]K_c[/tex] for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.
[tex]K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}[/tex].
[tex]K_c = 2.25[/tex]. As a result,
[tex]\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25[/tex].
[tex](3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0[/tex].
The degree of this polynomial is three. Plot the equation [tex]y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9[/tex] on a graph and look for any zeros. There's only one zero at [tex]x \approx 0.337[/tex]. All three concentrations end up greater than zero.
Hence the equilibrium concentration of Y: [tex]0.337\;\text{mol}\cdot\text{dm}^{-3}[/tex].
