【高校化学】結合エネルギーと反応熱(センター実戦)
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生成熱を求める問題
$\text{NH}_3$(気) 1 mol 中の N-H 結合をすべて切断するのに必要なエネルギーは何 kJ か。ただし,H-H および N≡N の結合エネルギーはそれぞれ 436 kJ/mol,945 kJ/mol であり,$\text{NH}_3$ (気)の生成熱は次の熱化学方程式で表されるものとする。(センター2017本試・改)
$\cfrac{\enspace3\enspace}{2}\text{H}_2$(気)$+\cfrac{\enspace1\enspace}{2}\text{N}_2$(気)$=\text{NH}_3$(気)$+46$ kJ
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エネルギーの出入り
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連立方程式をつくる
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$2\text{H}$(気)$=\text{H}_2$(気)$+436$ kJ ・・・①
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同じように N の式をつくると
$2\text{N}$(気)$=\text{N}_2$(気)$+945$ kJ ・・・②
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$2\text{H}$(気)$=\text{H}_2$(気)$+436$ kJ ・・・①
$2\text{N}$(気)$=\text{N}_2$(気)$+945$ kJ ・・・②
$\cfrac{\enspace3\enspace}{2}\text{H}_2$(気)$+\cfrac{\enspace1\enspace}{2}\text{N}_2$(気)$=\text{NH}_3$(気)$+46$ kJ ・・・③
式を連立して不要なものを消去
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①×$\cfrac{\space3\space}{2}$
$3\text{H}$(気)$=\cfrac{\space3\space}{2}\text{H}_2$(気)$+654$ kJ ・・・①’
②×$\cfrac{\space1\space}{2}$
$\text{N}$(気)$=\cfrac{\space1\space}{2}\text{N}_2$(気)$+472.5$ kJ ・・・②’
①’+②’+③
$3\text{H}$(気)$+\text{N}$(気)$+\cfrac{\space3\space}{2}\text{H}_2$(気)$+\cfrac{\space1\space}{2}\text{N}_2$(気)$=\cfrac{\space3\space}{2}\text{H}_2$(気)$+\cfrac{\space1\space}{2}\text{N}_2$(気)$+\text{NH}_3$(気)$+1172.5$ kJ
左右の同じ項を消去すると
$3\text{H}$(気)$+\text{N}$(気)$=\text{NH}_3$(気)$+1172.5$ kJ
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$Q=436\times\cfrac{\space3\space}{2}+945\times\cfrac{\space1\space}{2}+46=1170$ kJ
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SNSでシェア