②2CO(g)+O2(g)= 2CO 2(g)△H = _ 556.0 kj？ Mol-1
③H2O(l)=H2O(g)△H=+44.0kJ？ Mol-1
According to the thermochemical equation and Gass law ①-②-③×4, we can get ch3oh (l)+O2 (g) = co (g)+2h2o (l) △ h = _ 447.8kj? mol- 1；
So the answer is: CH3OH(l)+O2(g)=CO? (g)+2H2O(l)△H=_447.8？ kJ？ mol- 1；
(2) The methanol fuel cell is an experimental device for preparing iron hydroxide by electrolysis. After electrifying, a large number of white precipitates were produced in the solution, which did not change color for a long time, indicating that ferrous hydroxide was generated near electrode B, so electrode A was the anode of the electrolyzer and electrode B was the cathode of the electrolyzer. Power supply A is the positive electrode, and the electrode reacts with oxygen to obtain electrons for reduction reaction, with O2+4e-+2h2o = 4oh-; Electrode a connected to the positive electrode is iron. The purpose is that iron as anode loses electrons to generate ferrous ions, which react with hydroxide ions generated by cathode to generate white precipitated ferrous hydroxide. B is the negative electrode, and the electrode reaction connected with electrode B is 2h ++ 2e-= H2↑； ↑;
So, the answer is: positive, O2+4e-+2H2O=4OH-, Fe, 2H++2? E-= H2 = (or 2h2o+2e-= H2+2oh-);
③ ① carbon monoxide (g)+H2O(g)? CO2(g)+H2(g), the gas volume is unchanged before and after the reaction, the initial amount is reduced by half, and the equilibrium is static. When the concentration of carbon dioxide reaches equilibrium, it is 0.8 mol/L. Therefore, according to the chart data, it can be seen that the temperature rises and the equilibrium state moves in the opposite direction, indicating that the positive reaction is an exothermic reaction.
So, the answer is: release;
② Carbon monoxide (g) +H2O (g)? Carbon dioxide (g) +H2 (g)
Initial quantity (mole) 2? 4 ? 0 0
Variation (mole) 1.6 1.6? 1.6 1.6
Equilibrium quantity (mole) 0.4? 2.4 1.6? 1.6
In the experiment 1, v (H2) =1.6mol 2min = 0.16mol/(l? min)；
So, the answer is: 0. 16mol/(L? min)；
③ The equilibrium constant at ③900°C is calculated as follows
Carbon monoxide +H2O? Carbon dioxide (g) +H2 (g)
Starting amount (mol) 1 2? 00
Variation (mole) 0.4 0.4 0.4? 0.4
Equilibrium quantity (mole) 0.6? 1.6 0.4? 0.4
k = c(CO2)c(H2)c(CO)c(H2O)= 0.4mol 2l×0.4mol 2l 0.6mol 2l× 1.6mol 2l = 0. 17
According to the amounts of carbon monoxide (g), H2O (g), carbon dioxide (g) and H2 (g), 0.8 mol, 1.0 mol, 0.6 mol and 0.8 mol were added respectively.
QC = c(CO2)c(H2)c(CO)c(H2O)= 0.6mol 2 l×0.8mol 2 l 0.8mol 2 l× 1mol 2 l = 0.6 > K
The reaction proceeds in the opposite direction, v (positive) < v (negative).
So the answer is:
④ Compared with experiment 2, experiment 3 reached the same equilibrium state, the gas volume remained unchanged before and after the reaction, and the reaction rate increased, indicating that it was consistent to increase the pressure or add the catalyst, and the changing condition might be the use of the catalyst; Increase the pressure;
So, the answer is: use a catalyst; Increase the pressure;