英语周报牛津2022高一答案
19.(12分)(1)4NO2+4N2O4+3O2+6H2O=12HNO(2分)(2)-55.3(2分)(3)6%(2分)2945.3(2分)2分)③H(2分)【解析】(1)NO2和N2O4以物质的量之比1:1与O2和H2O恰好完全反应,化学方程式为4NO2+4N2O4+3O2+6H2O=12HNO3。(2)根据盖斯定律,I×2-Ⅱ得,2NO2(g)N2O4(g)△H=-4.4kJ·mol153.1kJ·mol-1=-55.3kJ·mol-1(3)①设起始时NO2(g)的物质的量为nmol,反应的物质的量为2xmol,根据三段式2NO2(g)=N2O4(g)起始(mol):n0反应(mol):2x平衡(mol):n-2xn2x+x97,解得x=3,NO2的转化率为则2m×100%=6%;平衡时n(NO2)=(n-2×n mmo=47nnoI, n100m,则p(NO2)=97P(NO2)×971Pa,p(N2O)=7×97kPa,Kp=p(N2O)2945.3kPa②B点压强大于E点,压强增大,化学反应速率加快,则v(B)>v(E)。③t2时刻移动活塞压强迅速增大,说明针筒内体积缩小,保持活塞位置木变后,平衡正向移动,混合气氵的物质的量逐渐减小,根据M=m可知,E、F、G、H四点中对应气体的平均相对分子质量最大的点氵为H
英语周报牛津2022高一答案
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英语周报牛津2022高一答案
19.(12分)(1)4NO2+4N2O4+3O2+6H2O=12HNO(2分)(2)-55.3(2分)(3)6%(2分)2945.3(2分)2分)③H(2分)【解析】(1)NO2和N2O4以物質的量之比1:1與O2和H2O恰好完全反應,化學方程式為4NO2+4N2O4+3O2+6H2O=12HNO3。(2)根據蓋斯定律,I×2-Ⅱ得,2NO2(g)N2O4(g)△H=-4.4kJ·mol153.1kJ·mol-1=-55.3kJ·mol-1(3)①設起始時NO2(g)的物質的量為nmol,反應的物質的量為2xmol,根據三段式2NO2(g)=N2O4(g)起始(mol):n0反應(mol):2x平衡(mol):n-2xn2x+x97,解得x=3,NO2的轉化率為則2m×100%=6%;平衡時n(NO2)=(n-2×n mmo=47nnoI, n100m,則p(NO2)=97P(NO2)×971Pa,p(N2O)=7×97kPa,Kp=p(N2O)2945.3kPa②B點壓強大於E點,壓強增大,化學反應速率加快,則v(B)>v(E)。③t2時刻移動活塞壓強迅速增大,說明針筒內體積縮小,保持活塞位置木變後,平衡正向移動,混合氣氵的物質的量逐漸減小,根據M=m可知,E、F、G、H四點中對應氣體的平均相對分子質量最大的點氵為H
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