题目

已知：N2(g)+3H2(g) 2NH3(g)；△H=-92.4 kJ/mol，下列结论正确的是 A．在密闭容器中加入1 molN2(g)和3 molH2(g)充分反应放热92.4 kJ B．N2(g)+3H2(g)  2NH3(l)；△H=-Q kJ/mol ，则Q＞92.4 C．增大压强，平衡向右移动，平衡常数增大 D．若一定条件下反应达到平衡，N2的转化率为20%，则H2的转化率一定为60% 答案：【答案】B 【解析】 试题分析：反应是可逆反应1 molN2(g)和3 molH2(g)不可能完全反应，放热不会达到92.4kJ，A错；B中反应同时包含了气态氨气液化成液态氨气放热过程，故B放出热量大于92.4kJ，B对；平衡常数与压强无关，C错；转化率与化学计量数无关，D错；选B. 考点：化学平衡与热化学结合。She told us that her brother ______ the Army for 3 years. A. had joined???????? B. had been in???? C. joined??????? D. has been in