题目内容
化学反应N2+3H2=2NH3的能量变化如下图所示,该反应的热化学方程式是
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A.N2(g)+3H2(g) === 2NH3(1) △H=2(a-b-c)kJ·mol-1
B.N2(g)+3H2(g) === 2NH3(g) △H=2(b-a)kJ·mol-1
C.N2(g)+H2(g) === NH3(1) △H=(b+c-a)kJ·mol-1
D.N2(g)+H2(g) === NH3(g) △H=(a+b)kJ·mol-1
B.N2(g)+3H2(g) === 2NH3(g) △H=2(b-a)kJ·mol-1
C.N2(g)+H2(g) === NH3(1) △H=(b+c-a)kJ·mol-1
D.N2(g)+H2(g) === NH3(g) △H=(a+b)kJ·mol-1
A
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| A、N2(g)+H2(g)→NH3(1)-46 kJ | B、N2(g)+H2(g)→NH3(g)-454 kJ | C、N2(g)+3 H2(g)→2 NH3(g)+92 kJ | D、N2(g)+3 H2(g)→2 NH3(1)+431.3 kJ |
| A、N2(g)+3H2(g)?2NH3(l);△H=2(a-b-c)kJ?mol-1 | ||||
| B、N2(g)+3H2(g)?2NH3(g);△H=2(b-a)kJ?mol-1 | ||||
C、
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D、
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