{"id":581,"date":"2016-01-04T15:47:24","date_gmt":"2016-01-04T15:47:24","guid":{"rendered":"https:\/\/courses.candelalearning.com\/chemistry111labs1x1xmaster\/?post_type=chapter&#038;p=581"},"modified":"2016-01-07T01:12:22","modified_gmt":"2016-01-07T01:12:22","slug":"lab-10-introduction","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/chapter\/lab-10-introduction\/","title":{"raw":"Lab 10 Introduction","rendered":"Lab 10 Introduction"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n\t<li>Use Hess\u2019 Law to calculate the enthalpy for a reaction.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2><strong>Introduction<\/strong><strong>\u00a0<\/strong><\/h2>\r\n[caption id=\"attachment_582\" align=\"alignright\" width=\"249\"]<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1314\/2016\/01\/03204619\/lab10box.png\"><img class=\"wp-image-582 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1314\/2016\/01\/03204619\/lab10box.png\" alt=\"Graphical representation of Hess's law The net reaction here is A being converted into D, and the change in enthalpy for that reaction is \u0394H. However, we can see that the net reaction is a result of A being converted into B, which is then converted into C, which is finally converted into D. By Hess's law, the net change in enthalpy of the overall reaction is equal to the sum of the changes in enthalpy for each intermediate transformation: \u0394H = \u0394H1+\u0394H2+\u0394H3. From Boundless.com.\" width=\"249\" height=\"216\" \/><\/a> Hess's Law. Source: <a href=\"https:\/\/www.boundless.com\/chemistry\/textbooks\/boundless-chemistry-textbook\/thermochemistry-6\/standard-enthalpy-of-formation-and-reaction-61\/hess-s-law-287-1534\/\">Boundless<\/a>. CC-BY-SA 3.0.[\/caption]\r\n\r\nHess\u2019 Law states that a reaction will have the same enthalpy change regardless of whether the reaction occurs in a single step or in multiple. This is beneficial in that we can use step reactions to find the enthalpy of a target reaction. Consider the image to the right. The energy change from A \u2192 D is the same as the sum of the changes from A \u2192 B, B \u2192 C, and C\u2192 D. This concept is occasionally used in industry to allow a reaction that is highly endothermic or slow to occur through stepwise reactions that are easier (or cheaper) to manipulate.\r\n\r\nWhen working a Hess\u2019 Law problem, you need to follow two main rules:\r\n<ol>\r\n\t<li>If you reverse the direction of a reaction, the sign of the \u2206H changes.<\/li>\r\n\t<li>If you multiply or divide the coefficients of a reaction, you must also multiply or divide the \u2206H value.<\/li>\r\n<\/ol>\r\nThese rules are applied to a set of steps in solving Hess\u2019 Law problems.\r\n<p style=\"padding-left: 30px;\"><strong>1. Evaluate<\/strong> the <strong>Target Reaction.\u00a0<\/strong>This reaction is the one you want to solve for and will be what you manipulate the step reactions for.<\/p>\r\n<p style=\"padding-left: 30px;\"><strong>2. Look <\/strong>over the <strong>Step Reactions<\/strong>.\u00a0These are the reactions you want to eventually \u201cadd up\u201d to get the target. You may wish to highlight or underline substances that are specifically in the Target Reaction. You may also wish you indicate substances that appear only one time throughout the Step Reactions. These will not be cancelled and will help you line up your reactions faster.<\/p>\r\n<p style=\"padding-left: 30px;\"><strong>3. Line up<\/strong> the <strong>Step Reactions<\/strong>.\u00a0You will need to rewrite the step reactions so that the reactants in the target reaction are listed on the left and products in the target reaction are on the right of the arrow. It is helpful if the arrows for all Step Reactions line up. Also be aware that some reactants or products may appear more than once and they will cancel (similar to spectator ions) when adding the reactions.<\/p>\r\n<p style=\"padding-left: 30px;\"><strong>4. Manipulate<\/strong> the <strong>Step Reactions<\/strong>.\u00a0Once your Step Reactions are lined up, you may need to multiply or divide the reactions to guarantee the correct number of each substance from the target reaction is present.<\/p>\r\n<p style=\"padding-left: 30px;\"><strong>5. Add<\/strong> the <strong>Step Reactions<\/strong>.\u00a0This will make sure that the sum of the step reactions will give the target reaction.<\/p>\r\n<p style=\"padding-left: 30px;\"><strong>6. Add <\/strong>the <strong><strong>Enthalpies.\u00a0<\/strong><\/strong>Once you have the ensured the Step Reactions add together to give the Target Reaction, Add the enthalpies of the reactions to give the enthalpy of the target reaction.<\/p>\r\nConsider the following example where we want to calculate the \u0394H values for the following reaction given the subsequent steps and enthalpy values.\r\n<ol>\r\n\t<li>Target Reaction:<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">C(s graphite) \u2192 C(s diamond) \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = ?<\/p>\r\n\r\n<ol start=\"2\">\r\n\t<li style=\"text-align: left;\">Step Reactions:<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">C(s graphite) \u2192 CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u0394H = -393.41 kJ<\/p>\r\n<p style=\"text-align: center;\">2C(s diamond) \u2192 2CO<sub>2<\/sub> (g) + 2O<sub>2<\/sub> (g) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u0394H = -790.80 kJ<\/p>\r\nWe know from the Target Reaction we want graphite on the left. It is on the left in the first Step Reaction, so we will probably leave this reaction alone and rewrite it as is. We also know that we want diamond on the right so we will need to reverse the second reaction to get diamond to the product side. We will need to reverse the sign of the second reaction. Lining up the Step Reactions gives:\r\n<p style=\"text-align: center;\">C(s graphite) \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2192 \u00a0CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = -393.41 kJ<\/p>\r\n<p style=\"text-align: center;\">2CO<sub>2<\/sub> (g) + 2O<sub>2<\/sub> (g) \u00a0\u00a0\u00a0\u00a0 \u2192 \u00a02C(s diamond)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = <span style=\"color: #ff0000;\">+<\/span>790.80 kJ<\/p>\r\nWe can tell that we will have too many diamond particles by the coefficients in the second Step Reaction. We will also not be able to cancel the carbon dioxide or oxygen because the coefficients are not the same in the first and second Target Reaction. Therefore we need to divide the WHOLE second reaction AND the enthalpy value by 2:\r\n<p style=\"text-align: center;\">C(s graphite) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2192 CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u00a0 \u00a0 \u00a0\u0394H = -393.41 kJ<\/p>\r\n<p style=\"text-align: center;\"><span style=\"color: #ff0000;\">1<\/span>CO<sub>2<\/sub> (g) + <span style=\"color: #ff0000;\">1<\/span>O2 (g) \u00a0 \u00a0\u2192 <span style=\"color: #ff0000;\">1<\/span>C(s diamond)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = <span style=\"color: #ff0000;\">395.4<\/span> kJ<\/p>\r\nAdding these reactions gives:\r\n<p style=\"text-align: center;\">C(s graphite) + CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u2192 CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) + C(s diamond)<\/p>\r\nNotice there are equal numbers of carbon dioxide and oxygen on the left AND right of the equation. Any substance on both sides will cancel. This gives:\r\n<p style=\"text-align: center;\">C(s graphite) \u2192 C(s diamond)<\/p>\r\nWhich is our Target Reaction. Therefore we can add the adjusted enthalpies to get the enthalpy of the target reaction.\r\n<p style=\"text-align: center;\">-393.41 kJ + 395.4 kJ =1.99 kJ<\/p>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Use Hess\u2019 Law to calculate the enthalpy for a reaction.<\/li>\n<\/ul>\n<\/div>\n<h2><strong>Introduction<\/strong><strong>\u00a0<\/strong><\/h2>\n<div id=\"attachment_582\" style=\"width: 259px\" class=\"wp-caption alignright\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1314\/2016\/01\/03204619\/lab10box.png\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-582\" class=\"wp-image-582 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1314\/2016\/01\/03204619\/lab10box.png\" alt=\"Graphical representation of Hess's law The net reaction here is A being converted into D, and the change in enthalpy for that reaction is \u0394H. However, we can see that the net reaction is a result of A being converted into B, which is then converted into C, which is finally converted into D. By Hess's law, the net change in enthalpy of the overall reaction is equal to the sum of the changes in enthalpy for each intermediate transformation: \u0394H = \u0394H1+\u0394H2+\u0394H3. From Boundless.com.\" width=\"249\" height=\"216\" \/><\/a><\/p>\n<p id=\"caption-attachment-582\" class=\"wp-caption-text\">Hess&#8217;s Law. Source: <a href=\"https:\/\/www.boundless.com\/chemistry\/textbooks\/boundless-chemistry-textbook\/thermochemistry-6\/standard-enthalpy-of-formation-and-reaction-61\/hess-s-law-287-1534\/\">Boundless<\/a>. CC-BY-SA 3.0.<\/p>\n<\/div>\n<p>Hess\u2019 Law states that a reaction will have the same enthalpy change regardless of whether the reaction occurs in a single step or in multiple. This is beneficial in that we can use step reactions to find the enthalpy of a target reaction. Consider the image to the right. The energy change from A \u2192 D is the same as the sum of the changes from A \u2192 B, B \u2192 C, and C\u2192 D. This concept is occasionally used in industry to allow a reaction that is highly endothermic or slow to occur through stepwise reactions that are easier (or cheaper) to manipulate.<\/p>\n<p>When working a Hess\u2019 Law problem, you need to follow two main rules:<\/p>\n<ol>\n<li>If you reverse the direction of a reaction, the sign of the \u2206H changes.<\/li>\n<li>If you multiply or divide the coefficients of a reaction, you must also multiply or divide the \u2206H value.<\/li>\n<\/ol>\n<p>These rules are applied to a set of steps in solving Hess\u2019 Law problems.<\/p>\n<p style=\"padding-left: 30px;\"><strong>1. Evaluate<\/strong> the <strong>Target Reaction.\u00a0<\/strong>This reaction is the one you want to solve for and will be what you manipulate the step reactions for.<\/p>\n<p style=\"padding-left: 30px;\"><strong>2. Look <\/strong>over the <strong>Step Reactions<\/strong>.\u00a0These are the reactions you want to eventually \u201cadd up\u201d to get the target. You may wish to highlight or underline substances that are specifically in the Target Reaction. You may also wish you indicate substances that appear only one time throughout the Step Reactions. These will not be cancelled and will help you line up your reactions faster.<\/p>\n<p style=\"padding-left: 30px;\"><strong>3. Line up<\/strong> the <strong>Step Reactions<\/strong>.\u00a0You will need to rewrite the step reactions so that the reactants in the target reaction are listed on the left and products in the target reaction are on the right of the arrow. It is helpful if the arrows for all Step Reactions line up. Also be aware that some reactants or products may appear more than once and they will cancel (similar to spectator ions) when adding the reactions.<\/p>\n<p style=\"padding-left: 30px;\"><strong>4. Manipulate<\/strong> the <strong>Step Reactions<\/strong>.\u00a0Once your Step Reactions are lined up, you may need to multiply or divide the reactions to guarantee the correct number of each substance from the target reaction is present.<\/p>\n<p style=\"padding-left: 30px;\"><strong>5. Add<\/strong> the <strong>Step Reactions<\/strong>.\u00a0This will make sure that the sum of the step reactions will give the target reaction.<\/p>\n<p style=\"padding-left: 30px;\"><strong>6. Add <\/strong>the <strong><strong>Enthalpies.\u00a0<\/strong><\/strong>Once you have the ensured the Step Reactions add together to give the Target Reaction, Add the enthalpies of the reactions to give the enthalpy of the target reaction.<\/p>\n<p>Consider the following example where we want to calculate the \u0394H values for the following reaction given the subsequent steps and enthalpy values.<\/p>\n<ol>\n<li>Target Reaction:<\/li>\n<\/ol>\n<p style=\"text-align: center;\">C(s graphite) \u2192 C(s diamond) \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = ?<\/p>\n<ol start=\"2\">\n<li style=\"text-align: left;\">Step Reactions:<\/li>\n<\/ol>\n<p style=\"text-align: center;\">C(s graphite) \u2192 CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u0394H = -393.41 kJ<\/p>\n<p style=\"text-align: center;\">2C(s diamond) \u2192 2CO<sub>2<\/sub> (g) + 2O<sub>2<\/sub> (g) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u0394H = -790.80 kJ<\/p>\n<p>We know from the Target Reaction we want graphite on the left. It is on the left in the first Step Reaction, so we will probably leave this reaction alone and rewrite it as is. We also know that we want diamond on the right so we will need to reverse the second reaction to get diamond to the product side. We will need to reverse the sign of the second reaction. Lining up the Step Reactions gives:<\/p>\n<p style=\"text-align: center;\">C(s graphite) \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2192 \u00a0CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = -393.41 kJ<\/p>\n<p style=\"text-align: center;\">2CO<sub>2<\/sub> (g) + 2O<sub>2<\/sub> (g) \u00a0\u00a0\u00a0\u00a0 \u2192 \u00a02C(s diamond)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = <span style=\"color: #ff0000;\">+<\/span>790.80 kJ<\/p>\n<p>We can tell that we will have too many diamond particles by the coefficients in the second Step Reaction. We will also not be able to cancel the carbon dioxide or oxygen because the coefficients are not the same in the first and second Target Reaction. Therefore we need to divide the WHOLE second reaction AND the enthalpy value by 2:<\/p>\n<p style=\"text-align: center;\">C(s graphite) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2192 CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u00a0 \u00a0 \u00a0\u0394H = -393.41 kJ<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000;\">1<\/span>CO<sub>2<\/sub> (g) + <span style=\"color: #ff0000;\">1<\/span>O2 (g) \u00a0 \u00a0\u2192 <span style=\"color: #ff0000;\">1<\/span>C(s diamond)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394H = <span style=\"color: #ff0000;\">395.4<\/span> kJ<\/p>\n<p>Adding these reactions gives:<\/p>\n<p style=\"text-align: center;\">C(s graphite) + CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) \u2192 CO<sub>2<\/sub> (g) +O<sub>2<\/sub> (g) + C(s diamond)<\/p>\n<p>Notice there are equal numbers of carbon dioxide and oxygen on the left AND right of the equation. Any substance on both sides will cancel. This gives:<\/p>\n<p style=\"text-align: center;\">C(s graphite) \u2192 C(s diamond)<\/p>\n<p>Which is our Target Reaction. Therefore we can add the adjusted enthalpies to get the enthalpy of the target reaction.<\/p>\n<p style=\"text-align: center;\">-393.41 kJ + 395.4 kJ =1.99 kJ<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-581\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>College Chemistry 1. <strong>Authored by<\/strong>: Jessica Garber-Morales. <strong>Provided by<\/strong>: Tidewater Community College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.tcc.edu\/\">http:\/\/www.tcc.edu\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Hess&#039;s Law of Heat Summation. <strong>Provided by<\/strong>: CK-12. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/section\/17.15\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/section\/17.15\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em><\/li><li>Hess&#039;s Law problems. <strong>Authored by<\/strong>: CBender. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/goo.gl\/XNR1eU\">http:\/\/goo.gl\/XNR1eU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":74,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"College Chemistry 1\",\"author\":\"Jessica Garber-Morales\",\"organization\":\"Tidewater Community College\",\"url\":\"http:\/\/www.tcc.edu\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Hess\\'s Law of Heat Summation\",\"author\":\"\",\"organization\":\"CK-12\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/section\/17.15\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Hess\\'s Law problems\",\"author\":\"CBender\",\"organization\":\"\",\"url\":\"http:\/\/goo.gl\/XNR1eU\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-581","chapter","type-chapter","status-publish","hentry"],"part":68,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/pressbooks\/v2\/chapters\/581","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/wp\/v2\/users\/74"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/pressbooks\/v2\/chapters\/581\/revisions"}],"predecessor-version":[{"id":588,"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/pressbooks\/v2\/chapters\/581\/revisions\/588"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/pressbooks\/v2\/parts\/68"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/pressbooks\/v2\/chapters\/581\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/wp\/v2\/media?parent=581"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/pressbooks\/v2\/chapter-type?post=581"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/wp\/v2\/contributor?post=581"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chemistry1labs\/wp-json\/wp\/v2\/license?post=581"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}