Inverse reaction weather

(1)① As can be seen from the figure 1, the temperature T 1 reaches equilibrium first, so the temperature t 1 > t2. The higher the temperature, the lower the concentration of carbon dioxide, indicating that the temperature balance moves to the reverse reaction, so the positive and negative reactions should be exothermic, that is, △ h < 0.

So the answer is:

② As can be seen from the figure, 2s reaches equilibrium at T2 temperature, and the change of carbon dioxide concentration at equilibrium is 0. 1mol/L, so V (CO2) = 0.1mol/L2s = 0.05mol/(L? S), the rate ratio is equal to the stoichiometric ratio, so v (N2) =12v (CO2) =12x0.05mol/(l? s)=0.025mol/(L？ s)，

So the answer is: 0.025mol/(L? s)；

③ The larger the contact area, the faster the reaction speed and the shorter the time to reach equilibrium. The surface area of the catalyst is S 1 > S2, 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 temperature T 1, so the change curve of c(CO2) at T 1 and S2 is as follows:

So the answer is:

(4) A. After reaching equilibrium, the forward and reverse rates are equal and will not change. When t 1, v is the largest, and then it changes with the reaction rate, and it does not reach equilibrium, so a is wrong;

B, positive and negative reactions should be exothermic reaction, with the increase of reaction temperature, the chemical equilibrium constant decreases, after reaching equilibrium, the temperature remains the same, reaches the highest, the equilibrium constant remains the same, and is the minimum, and the image conforms to the reality, so B is correct.

C, after t 1, the amount of carbon dioxide and NO changes, and t 1 does not reach an equilibrium state, so C is wrong;

The mass fractions of D and NO are constant, and t 1 is always in equilibrium, so D is correct.

So the answer is: BD;

(2)① Known: Ⅰ. CH4 (g) +2NO2 (g) ═ N2 (g) +CO2 (g) +2H2O (g) △ H 1 =-867kJ/mol.

Ⅱ. 2NO2 (g)? N2O4 (g) △H2 =-56.9 kJ/mol

According to Gaith's law, I-II obtains CH4 (g)+N2O4 (g) = N2 (g)+CO2 (g)+2h2o (g), so △ h =-867 kJ/mol-(-56.9 kJ/mol) =-810.65438.

That is, CH4 (g)+N2O4 (g) = N2 (g)+CO2 (g)+2h2o (g), △H=-8 10. 1kJ/mol.

So the answer is: CH4 (g)+N2O4 (g) = N2 (g)+CO2 (g)+2h2o (g), △ h =-810.1kj/mol;

(2) As can be seen from the figure, water is introduced into the left chamber to generate oxygen and hydrogen ions, and the surface of catalyst A is oxidized to be a negative electrode, while carbon dioxide is introduced into the right chamber to generate HCOOH under acidic conditions. The electrode reaction formula is CO2+2h+2e-= HCOOH.

So the answer is: CO2+2h+2e-= HCOOH;

③ At room temperature, the pH of 0. 1mol/L HCOONa solution is 10, and HCOO-+H2O hydrolyzed by HCOO exists in the solution. HCOOH+OH-, so Kh= 10? 4× 10? 40. 1? 10? 4= 10-7, then the ionization constant ka of HCOOH = kwkh =10? 14 10? 7= 10-7,

So the answer is: 10-7.