{"id":9224,"date":"2024-08-06T20:41:10","date_gmt":"2024-08-06T23:41:10","guid":{"rendered":"https:\/\/xequematenem.com.br\/blog\/?p=9224"},"modified":"2024-08-06T20:41:39","modified_gmt":"2024-08-06T23:41:39","slug":"questao-50-enem-2011","status":"publish","type":"post","link":"https:\/\/xequematenem.com.br\/blog\/questao-50-enem-2011\/","title":{"rendered":"Quest\u00e3o 050 \u2013 ENEM 2011"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Um dos problemas dos combust\u00edveis que cont\u00eam carbono \u00e9 que sua queima produz di\u00f3xido de carbono. Portanto, uma caracter\u00edstica importante, ao se escolher um combust\u00edvel, \u00e9 analisar seu calor de combust\u00e3o (\u2206H<sup>o<\/sup><sub>C<\/sub>), definido como a energia liberada na queima completa de um mol de combust\u00edvel no estado padr\u00e3o. O quadro seguinte relaciona algumas subst\u00e2ncias que cont\u00eam carbono e seu \u2206H<sup>o<\/sup><sub>C<\/sub><\/p>\n\n\n\n<figure class=\"wp-block-image size-full is-resized\"><img fetchpriority=\"high\" decoding=\"async\" width=\"532\" height=\"254\" src=\"https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-362.png\" alt=\"\" class=\"wp-image-9225\" style=\"width:423px;height:auto\" title=\"\" srcset=\"https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-362.png 532w, https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-362-300x143.png 300w\" sizes=\"(max-width: 532px) 100vw, 532px\" \/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Neste contexto, <strong>qual dos combust\u00edveis, quando queimado completamente, libera mais di\u00f3xido de carbono no ambiente pela mesma quantidade de energia produzida?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A) Benzeno.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">B) Metano.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">C) Glicose.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">D) Octano.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E) Etanol.<\/p>\n\n\n\n<p class=\"has-text-align-center has-large-font-size wp-block-paragraph\"><strong>Solu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Considerando uma quantidade fixa e igual de energia para todas as subst\u00e2ncias, podemos estabelecer uma rela\u00e7\u00e3o entre a quantidade de CO\u2082 liberada e a energia produzida. Durante uma combust\u00e3o completa, s\u00e3o gerados CO\u2082 e H\u2082O. A quantidade em mol de CO\u2082 produzida est\u00e1 diretamente relacionada ao n\u00famero de \u00e1tomos de carbono presentes no combust\u00edvel, o que pode ser determinado atrav\u00e9s do balanceamento da rea\u00e7\u00e3o de combust\u00e3o.<\/p>\n\n\n\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-2c90304e wp-block-group-is-layout-flex\">\n<p class=\"wp-block-paragraph\"><strong>Benzeno: C<sub>6<\/sub>H<sub>6<\/sub> (l) \u2192 6 CO<sub>2<\/sub> (g)&nbsp; \u2206H<sup>o<\/sup><sub>C<\/sub> = \u2013 3 268 kJ<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">3 268 kJ de energia liberada &#8212;&#8212;&#8212;&#8211; 6 mol de CO<sub>2<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 kJ de energia liberada &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; x<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>x = 0,0018 mol de CO<sub>2<\/sub><\/strong><\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-2c90304e wp-block-group-is-layout-flex\">\n<p class=\"wp-block-paragraph\"><strong>Etanol: C<sub>2<\/sub>H<sub>5<\/sub>OH (l) \u2192 2 CO<sub>2<\/sub> (g)&nbsp;&nbsp; \u2206H<sup>o<\/sup><sub>C<\/sub> = \u2013 1 368 kJ<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 368 kJ de energia liberada &#8212;&#8212;&#8212;&#8211; 2 mol de CO<sub>2<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 kJ de energia liberada &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; y<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>y = 0,0014 mol de CO<sub>2<\/sub><\/strong><\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-2c90304e wp-block-group-is-layout-flex\">\n<p class=\"wp-block-paragraph\"><strong>Glicose: C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> (s) \u2192 6 CO<sub>2<\/sub> (g)&nbsp; &nbsp;\u2206H<sup>o<\/sup><sub>C<\/sub> =<\/strong> <strong>\u2013 2 808 kJ<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">2 808 kJ de energia liberada &#8212;&#8212;&#8212;&#8211; 6 mol de CO<sub>2<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 kJ de energia liberada &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; w<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>w = 0,0021 mol de CO<sub>2<\/sub><\/strong><\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-2c90304e wp-block-group-is-layout-flex\">\n<p class=\"wp-block-paragraph\"><strong>Metano: CH<sub>4<\/sub> (g) \u2192 CO<sub>2<\/sub> (g)&nbsp;&nbsp; \u2206H<sup>o<\/sup><sub>C<\/sub> = \u2013 890 kJ<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">890 kJ de energia liberada &#8212;&#8212;&#8212;&#8211; 1 mol de CO<sub>2<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 kJ de energia liberada &#8212;&#8212;&#8212;&#8212;&#8212; z<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>z = 0,0011 mol de CO<sub>2<\/sub><\/strong><\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-2c90304e wp-block-group-is-layout-flex\">\n<p class=\"wp-block-paragraph\"><strong>Octano: C<sub>8<\/sub>H<sub>18<\/sub> (l) \u2192 8 CO<sub>2<\/sub> (g)&nbsp;&nbsp; \u2206H<sup>o<\/sup><sub>C<\/sub> = \u2013 5 471 kJ<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">5 471 kJ de energia liberada &#8212;&#8212;&#8212;&#8211; 8 mol de CO<sub>2<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 kJ de energia liberada &#8212;&#8212;&#8212;&#8212;&#8212; k<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>k = 0,0014 mol de CO<sub>2<\/sub><\/strong><\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A glicose libera mais di\u00f3xido de carbono, considerando a mesma quantidade de energia produzida (1 kJ)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Alternativa C<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Um dos problemas dos combust\u00edveis que cont\u00eam carbono \u00e9 que sua queima produz di\u00f3xido de carbono. Portanto, uma caracter\u00edstica importante, ao se escolher um combust\u00edvel, \u00e9 analisar seu calor de combust\u00e3o (\u2206HoC), definido como a energia liberada na queima completa de um mol de combust\u00edvel no estado padr\u00e3o. O quadro seguinte relaciona algumas subst\u00e2ncias que [&hellip;]<\/p>\n","protected":false},"author":11,"featured_media":9225,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[17,28],"tags":[],"area-do-conhecimento":[116,120],"assunto":[],"class_list":["post-9224","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-questoes-do-enem","category-28","area-do-conhecimento-ciencias-da-natureza-e-suas-tecnologias","area-do-conhecimento-quimica"],"_links":{"self":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts\/9224","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/users\/11"}],"replies":[{"embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/comments?post=9224"}],"version-history":[{"count":2,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts\/9224\/revisions"}],"predecessor-version":[{"id":9227,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts\/9224\/revisions\/9227"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/media\/9225"}],"wp:attachment":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/media?parent=9224"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/categories?post=9224"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/tags?post=9224"},{"taxonomy":"area-do-conhecimento","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/area-do-conhecimento?post=9224"},{"taxonomy":"assunto","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/assunto?post=9224"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}