To determine the enthalpy change (\u0394rH\u00b0) for the reaction:<\/p>\n\n\n\n
CaC\u2082(s) + 2 H\u2082O(l) \u2192 Ca(OH)\u2082(s) + C\u2082H\u2082(g)<\/strong><\/p>\n\n\n\n We will use Hess’s Law<\/strong>, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.<\/p>\n\n\n\n We need to rearrange and combine the given reactions to reflect:<\/p>\n\n\n\n CaC\u2082(s) + 2 H\u2082O(l) \u2192 Ca(OH)\u2082(s) + C\u2082H\u2082(g)<\/strong><\/p>\n\n\n\n Already given: Ca(s) + \u00bd O\u2082 \u2192 CaO(s) \u2192 \u0394H = -635.5 kJ But we need Ca(OH)\u2082 on the product side, so we reverse this: C\u2082H\u2082(g) + 2.5 O\u2082 \u2192 2 CO\u2082 + H\u2082O \u2192 \u0394H = -1300 kJ 2 C(gr) + 2 O\u2082 \u2192 2 CO\u2082 \u2192 \u0394H = 2 \u00d7 (-393.5) = -787.0 kJ<\/strong><\/p>\n\n\n\n Use:<\/p>\n\n\n\n Add all:<\/p>\n\n\n\n \u0394rH\u00b0 = -62.8 + 700.7 + 1300 + (-787.0) \u0394rH\u00b0 = +1150.9 kJ\/mol<\/strong><\/p>\n\n\n\n To calculate the enthalpy change (\u0394rH\u00b0) for a chemical reaction that is not directly listed among standard thermochemical equations, we use Hess\u2019s Law<\/strong>. This law states that the total enthalpy change for a reaction is independent of the path taken and depends only on the initial and final states.<\/p>\n\n\n\n In this case, we are determining the enthalpy change for the reaction of calcium carbide (CaC\u2082) with water to form calcium hydroxide (Ca(OH)\u2082) and acetylene gas (C\u2082H\u2082). The approach is to break this overall reaction into known steps for which enthalpy values are provided.<\/p>\n\n\n\n First, we used the formation of CaC\u2082 from calcium metal and graphite. Then, we considered the reverse of calcium hydroxide formation to break it down into its elemental components. We also reversed the combustion of acetylene to reflect its formation from CO\u2082 and H\u2082O. Finally, we included the formation of CO\u2082 from graphite to close the cycle.<\/p>\n\n\n\n By summing the enthalpies of these adjusted reactions, Hess\u2019s Law allows us to derive the enthalpy change for the target reaction. The result is a positive enthalpy change of +1150.9 kJ\/mol<\/strong>, meaning the reaction is endothermic<\/strong>. This indicates that the process requires a significant input of energy to proceed, which is consistent with the nature of reactions that produce gaseous hydrocarbons like acetylene.<\/p>\n\n\n\n Acetylene gas C?H? can be produced by the reaction of calcium carbide CaC? with water according to the following balanced equation: CaC?(s) + 2 H?O(l)? Ca(OH)?(s) + C?H?(g) Determine the value of ?rH\u00b0 for this reaction using the thermochemical equations provided below: Ca(s) + 2 C(gr) ? CaC?(s) ?rH\u00b0 = -62.8 kJ mol?\u00b9 Ca(s) + […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-26718","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26718","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=26718"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26718\/revisions"}],"predecessor-version":[{"id":26720,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26718\/revisions\/26720"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26718"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26718"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26718"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Write out the given thermochemical equations<\/strong><\/h3>\n\n\n\n
\n
\u0394H\u00b0 = -62.8 kJ\/mol<\/li>\n\n\n\n
\u0394H\u00b0 = -635.5 kJ\/mol<\/li>\n\n\n\n
\u0394H\u00b0 = -65.2 kJ\/mol<\/li>\n\n\n\n
\u0394H\u00b0 = -2600 kJ\/mol
(Per mole of C\u2082H\u2082: divide by 2 \u2192 -1300 kJ\/mol)<\/li>\n\n\n\n
\u0394H\u00b0 = -393.5 kJ\/mol<\/li>\n<\/ol>\n\n\n\n
\n\n\n\nStep 2: Build a thermochemical cycle<\/strong><\/h3>\n\n\n\n
a) Formation of CaC\u2082 (from Ca and C):<\/h4>\n\n\n\n
Ca(s) + 2 C(gr) \u2192 CaC\u2082(s) \u2192 \u0394H = -62.8 kJ<\/p>\n\n\n\nb) Convert Ca(s) into Ca(OH)\u2082:<\/h4>\n\n\n\n
CaO(s) + H\u2082O(l) \u2192 Ca(OH)\u2082(s) \u2192 \u0394H = -65.2 kJ
Total for Ca(s) + H\u2082O \u2192 Ca(OH)\u2082(s):
\u0394H = -635.5 + (-65.2) = -700.7 kJ<\/strong><\/p>\n\n\n\n
Ca(OH)\u2082(s) \u2192 Ca(s) + \u00bd O\u2082 + H\u2082O(l) \u2192 \u0394H = +700.7 kJ<\/strong><\/p>\n\n\n\nc) Combustion of C\u2082H\u2082:<\/h4>\n\n\n\n
We reverse this to form C\u2082H\u2082:
2 CO\u2082 + H\u2082O \u2192 C\u2082H\u2082 + 2.5 O\u2082 \u2192 \u0394H = +1300 kJ<\/strong><\/p>\n\n\n\nd) CO\u2082 from graphite:<\/h4>\n\n\n\n
\n\n\n\nStep 3: Apply Hess’s Law<\/strong><\/h3>\n\n\n\n
\n
\u0394rH\u00b0 = +1150.9 kJ<\/strong><\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
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