普化习题答案

1.答案（1-2-3+4-）2.答案（1c2d3a4d5abd6ad7d8d）3.答案（1）燃烧前后系统的温度（2）水的质量和比热（3）弹式3.8g?1?2820kj?mol?30%?17.8kjqp?3(?27.6)?(?58.6)?2(38.1)??16.7kj?mol?nh4nh3(g)＋h2s(g)估算1mol反应的△h=-(-1206.92)+(-635.09)化焓乘以双键个数的差值.共轭双键的离域作用是分子变的稳定)1.答案（1-2-3-4-5-6+7+8+）-1-14.答案dceab）原因是气体熵液体固体大的rm(fm(298.15k,fe3o4)-4?fgmfgm(298.15k,fe3o4)=3[(-77.7)+rgm(298.15k)?(?236.7)?(?237.129)【篇二：普通化学习题参考答案】10.封闭系统经过一循环过程后，其下列哪组参数是?13.298k时，下列各物质的?fhm等于零的是哪个？(a)b.ch4(g)?2o2(g)===co2ch4(g)?3o2(g)===co(g)?2h2ocaco3(s)b.解1）w????b(g)rt??1?8.314?10?3?373.15??3.10kj.mol?1b（2）?u?qv?qp???b(g)rt?40.63?1?8.314?10?3?373.15?37.53kj.b??rhm?qp?qv???b(g)rt??82.7?(2?bw????(b,g)rt??1?8.314?10?3?291??2.42kj.mob?u?qv?qp???b,grt??15b列的各反应条件，计算发生下列变化时，各自?u与?h之间（1）?u??h?????b,grt??2?(1?1)?8.314?10?bb???rhm(298.15k)???b?fhm?4?(?92.3bb=-429.86-(4-4-1)?8.314?10?3?298.15?体????b,grt??(4?4?1)?8.314?10?3?298.15?2.479kj.b1,,??1fe3o4(s)?co(g)===3feo(s)?co2(g)，?rhm3=38.1kj?molm,4??3?r??m,1??rhm,2?2?rhm,3???16.7kj.mol?16b【篇三：普通化学习题参考答案】10.封闭系统经过一循环过程后，其下列哪组参数是?13.298k时，下列各物质的?fhm等于零的是哪个？(a)b.ch4(g)?2o2(g)===co2ch4(g)?3o2(g)===co(g)?2h2ocaco3(s)b.解1）w????b(g)rt??1?8.314?10?3?373.15??3.10kj.mol?1b（2）?u?qv?qp???b(g)rt?40.63?1?8.314?10?b??rhm?qp?qv???b(g)rt??82.7?(2?bw????(b,g)rt??1?8.314?10?3?291??2.42kj.mob?u?qv?qp???b,grt??15b列的各反应条件，计算发生下列变化时，各自?u与?h之间（1）?u??h?????b,grt??2?(1?1)?8.314?10?bb298.15k进行???rhm(298.15k)???b?fhm?4?(?92.3bb=-429.86-(4-4-1)?8.314?10?3?298.15?体????b,grt??(4?4?1)?8.314?10?3?298.15?2.479kj.bc6h12o6(s)?6o2(g)===6co2(g)?6h2o(l)?1?38.1