20 14 misty weather

( 1) T 2； When the temperature is T 1, the equilibrium concentration of carbon dioxide is low, which indicates that the temperature equilibrium moves to reverse reaction. According to the influence law of temperature on chemical equilibrium, the reaction is exothermic, △ h < 0; (2) According to the given image, 2s reaches equilibrium at T 2 temperature, and the carbon dioxide concentration becomes 0.1mol/L. According to the formula v=△c/△t, v(CO 2 )=0.05mol/(L? S), using a rate ratio equal to the stoichiometric ratio, v(N 2 )=0.025mol/(L? s)； (3) The larger the surface area of solid catalyst, the faster the reaction rate and the shorter the time to reach equilibrium. The surface area of the catalyst is S 1 >S 2, which takes a long time to reach equilibrium, but the catalyst does not affect the equilibrium movement. The concentration of carbon dioxide at equilibrium is the same as that at T 1, so c(CO 2) reaches an equilibrium process at T 1 and S 2. (4) A. After the reversible reaction reaches equilibrium, the positive and negative rates are equal and will not change. From Figure A, it can be seen that V is the positive maximum of t 1 time, and then it changes with the reaction rate, and it is wrong to fail to reach equilibrium. B, positive and negative reactions should be exothermic. The chemical equilibrium constant decreases with the increase of temperature. After reaching equilibrium, the temperature and equilibrium constant remain unchanged, which is consistent with the image. Correct, the amount of carbon dioxide has not changed after C and t 1, but t 1 has not reached an equilibrium state. Mistakes; The mass fractions of D and NO are constant, and t 1 is always in equilibrium, which is correct. Select bd.