(1)该燃料电池的正极发生还原反应,氧气在正极放电生成氢氧根离子,即6H2O+3O2+12e-=12OH-或2H2O+O2+4e-=4OH-,
故答案为:6H2O+3O2+12e-=12OH-或2H2O+O2+4e-=4OH-;
(2)已知:①2H2(g)+CO(g)?CH3OH(g)△H=a kJ?mol-1
②2CH3OH(g)?CH3OCH3(g)+H2O(g)△H=b kJ?mol-1
③CO(g)+H2O(g)?CO2(g)+H2(g)△H=ckJ?mol-1
由盖斯定律②+③+①×2得到3H2(g)+3CO(g)?CH3OCH3(g)+CO2(g)的△H=(2a+b+c)kJ?mol-1,
故答案为:(2a+b+c);
(3)①QC=
C(CH3OCH3)?c(H2O) |
c2(CH3OH) |
==2.25<400,QC小于平衡常数,反应正向进行尚未达到平衡,所以v(正)>v(逆),
故答案为:>;该时刻的浓度商QC=2.25<400,QC小于平衡常数,反应正向进行尚未达到平衡,所以v(正)>v(逆);
②由2CH3OH(g)═CH3OCH3(g)+H2O(g)和表格中的数据可知,开始c(CH3OH)=(0.64+0.5×2)mol/L=1.64mol/L,设达平衡时CH3OCH3的物质的量为x,则
2CH3OH(g)═CH3OCH3(g)+H2O(g)
起始量(mol/L) 1.6 0 0
变化量(mol/L) 2x x x
平衡量(mol/L)1.64-2x x x
K=
C(CH3OCH3)?c(H2O) |
c2(CH3OH) |
==400,解得X=0.8,
α(CH3OH)=×100%≈97.6%,
故答案为:97.6%.