强酸与强碱在稀溶液中和热可表示为H^+(ap)+OH^-(ap)=H2O(l) 焓变=-57.3kj/mol已知:接上CH3COOH+NaOH=CH3COONa+H2O 焓变=-Q1kj/mol1/2H2SO4(浓)+NaOH=1/2Na2SO4+H2O 焓变=-Q2kj/molHNO3+NaOH=NaNO3H2O 焓变=-Q3kj/mol上述反应
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![强酸与强碱在稀溶液中和热可表示为H^+(ap)+OH^-(ap)=H2O(l) 焓变=-57.3kj/mol已知:接上CH3COOH+NaOH=CH3COONa+H2O 焓变=-Q1kj/mol1/2H2SO4(浓)+NaOH=1/2Na2SO4+H2O 焓变=-Q2kj/molHNO3+NaOH=NaNO3H2O 焓变=-Q3kj/mol上述反应](/uploads/image/z/2368438-70-8.jpg?t=%E5%BC%BA%E9%85%B8%E4%B8%8E%E5%BC%BA%E7%A2%B1%E5%9C%A8%E7%A8%80%E6%BA%B6%E6%B6%B2%E4%B8%AD%E5%92%8C%E7%83%AD%E5%8F%AF%E8%A1%A8%E7%A4%BA%E4%B8%BAH%5E%2B%EF%BC%88ap%29%2BOH%5E-%28ap%29%3DH2O%28l%29+%E7%84%93%E5%8F%98%3D-57.3kj%2Fmol%E5%B7%B2%E7%9F%A5%EF%BC%9A%E6%8E%A5%E4%B8%8ACH3COOH%2BNaOH%3DCH3COONa%2BH2O+%E7%84%93%E5%8F%98%3D-Q1kj%2Fmol1%2F2H2SO4%EF%BC%88%E6%B5%93%EF%BC%89%2BNaOH%3D1%2F2Na2SO4%2BH2O+%E7%84%93%E5%8F%98%3D-Q2kj%2FmolHNO3%2BNaOH%3DNaNO3H2O+%E7%84%93%E5%8F%98%3D-Q3kj%2Fmol%E4%B8%8A%E8%BF%B0%E5%8F%8D%E5%BA%94)
强酸与强碱在稀溶液中和热可表示为H^+(ap)+OH^-(ap)=H2O(l) 焓变=-57.3kj/mol已知:接上CH3COOH+NaOH=CH3COONa+H2O 焓变=-Q1kj/mol1/2H2SO4(浓)+NaOH=1/2Na2SO4+H2O 焓变=-Q2kj/molHNO3+NaOH=NaNO3H2O 焓变=-Q3kj/mol上述反应
强酸与强碱在稀溶液中和热可表示为H^+(ap)+OH^-(ap)=H2O(l) 焓变=-57.3kj/mol已知:
接上CH3COOH+NaOH=CH3COONa+H2O 焓变=-Q1kj/mol
1/2H2SO4(浓)+NaOH=1/2Na2SO4+H2O 焓变=-Q2kj/mol
HNO3+NaOH=NaNO3H2O 焓变=-Q3kj/mol
上述反应均在溶液中进行Q1.Q2.Q3的关系正确的是
A.Q1=Q2=Q3 B.Q2>Q1>Q3 C.Q2>Q3>Q1 Q2=Q3>Q1
应该选什么 具体分析
强酸与强碱在稀溶液中和热可表示为H^+(ap)+OH^-(ap)=H2O(l) 焓变=-57.3kj/mol已知:接上CH3COOH+NaOH=CH3COONa+H2O 焓变=-Q1kj/mol1/2H2SO4(浓)+NaOH=1/2Na2SO4+H2O 焓变=-Q2kj/molHNO3+NaOH=NaNO3H2O 焓变=-Q3kj/mol上述反应
选C
CH3COOH是弱酸,电离过程要吸热,所以Q1最小;浓硫酸溶于水要放热,所以Q2最大,而HNO3的Q3正好是中和热的数值,介于Q1和Q3之间,题目中的Q前面有负号,所以不必考虑负号的影响,所以选C.Q2>Q3>Q1