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When an equilibrium mixture of gaseous, colorless $\ce{N2O4}$ and brown $\ce{NO2}$ is warmed at constant volume, which of the following is correct?

  1. The density remains constant.
  2. The degree of dissociation decreases.
  3. The average molar mass increases.
  4. The pressure decreases.
  5. The color becomes lighter.

The correct answer is (1). But all the other answers except (4) seem reasonable to me.

The equilibrium we are looking at is: $\ce{N2O4<=>2NO2}$

Upon heating at constant volume, the pressure will increase, and so the equilibrium will shift to the left, which means that there is less dissociation of $\ce{N2O4}$, the average molar mass will be closer to the molar mass of $\ce{N2O4}$, than before warming, thus the average molar mass will increase.

Finally, as the equilibrium shifts to the left, more colorless $\ce{N2O4}$ will be present, lightening the color of the gaseous mixture.

Obviously, the density will also remain constant, as density of gas = $\frac {\text{mass}}{\text{volume}}$ and since no mass is lost and we are at constant volume, the density remains constant.

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    $\begingroup$ Are you sure about the direction of the shift in equilibrium? True, increased pressure favors association and thus shifts it to the left; but increased temperature favors entropy and thus shifts it to the right. Which influence will prevail? $\endgroup$ – Ivan Neretin Sep 14 '15 at 17:47
  • $\begingroup$ What is the $\Delta H^\circ$ of the reaction? If the reaction has $\Delta_r H^\circ <0$, then increasing the temperature (adding heat) can cause the equilibrium to shift to the left. $\endgroup$ – Ben Norris Sep 14 '15 at 20:22
  • $\begingroup$ @IvanNeretin So, because we don't necessarily know if entropy or pressure wins out in the end, the only thing we do know is that no mass can escape, ergo, the density is constant? $\endgroup$ – coloratura Sep 14 '15 at 22:22

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