![K_{\text{c}} = \dfrac{\text{[NO$_{2}$]}^{2}}{\text{[N$_{2}$O$_{4}$]}} = \dfrac{(1.78 \times 10^{-2})^{2}}{\text{[N$_{2}$O$_{4}$]}} = 5.85 \times 10^{-3}\\\\\begin{array}{rcl}\\(1.78 \times 10^{-2})^{2}&=&\text{[N$_{2}$O$_{4}$]} \times 5.85 \times 10^{-3}\\\text{[N$_{2}$O$_4$]}&=& \dfrac{(1.78 \times 10^{-2})^{2}}{5.85 \times 10^{-3}}\\\\& = & \mathbf{5.41 \times 10^{-2}}\textbf{ mol/L}\\\end{array}\\](/tpl/images/0686/6303/751d3.png)
Consider the following equilibrium:
N204(9)
2NO2(9)
Ket
= 5.85 x 10-3
Whi...
Chemistry, 16.06.2020 01:57 Aliciaonfleek
Consider the following equilibrium:
N204(9)
2NO2(9)
Ket
= 5.85 x 10-3
Which statement about this system is true?
Options:
The equilibrium lies to the left.
The equilibrium lies to the right.
If the equilibrium concentration of No, is 1.78 x 10-2 m, the equilibrium concentration of
N204 is
Options:
3.04 M.
1.85 × 10^-6 M.
5.42 × 10^-2 M.
Answers: 3
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