摘要:已知在1×105 Pa.298 K条件下.2 mol氢气燃烧生成水蒸气放出484 kJ热量, 则下列热化学方程式正确的是 A.2 H2(g)+ O2(g)===2 H2O(g) △H=+484 kJ·mol-1 B.2 H2(g)+ O2(g)===2 H2O(l) △H=-484 kJ·mol-1 C.H2(g)+1/2 O2(g)====H2O(g) △H=+242 kJ·mol-1 D.H2O(g) == H2(g)+1/2 O2(g) △H=+242 kJ·mol-1

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