What is the enthalpy change, \u0394H, for this reaction? Show your work.<\/p>\n\n\n\n
The Correct answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct answer is: (-393.5 \\, \\text{kJ\/mol})<\/strong><\/mark><\/p>\n\n\n\n To calculate the enthalpy change (( \\Delta H )) for a chemical reaction, you generally use the formula:<\/p>\n\n\n\n [ where ( H ) represents the standard enthalpy of formation of the substances involved in the reaction. The enthalpy of formation (( \\Delta H_f^\\circ )) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.<\/p>\n\n\n\n Consider the reaction:<\/p>\n\n\n\n [ The enthalpy change (( \\Delta H )) for the combustion of carbon to form carbon dioxide is (-393.5 \\, \\text{kJ\/mol}). This negative sign indicates that the reaction is exothermic, meaning it releases heat to the surroundings.<\/p>\n\n\n\n The enthalpy change is an important concept in thermodynamics that reflects the energy changes during a chemical reaction. It provides insights into the feasibility of reactions, guiding chemists in predicting whether a reaction will occur spontaneously under given conditions. In the example above, the formation of carbon dioxide from carbon and oxygen is a highly exothermic reaction, which is a common characteristic of combustion processes. Understanding these changes helps in various applications, including energy production, environmental chemistry, and industrial processes. The negative ( \\Delta H ) value signifies that energy is released, making the reaction energetically favorable, a principle crucial in fields like combustion engineering and environmental science.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the enthalpy change, \u0394H, for this reaction? Show your work. The Correct answer and Explanation is : The correct answer is: (-393.5 \\, \\text{kJ\/mol}) To calculate the enthalpy change (( \\Delta H )) for a chemical reaction, you generally use the formula: [\\Delta H = \\sum H_{\\text{products}} – \\sum H_{\\text{reactants}}] where ( H […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-149907","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/149907","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=149907"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/149907\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=149907"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=149907"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=149907"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\Delta H = \\sum H_{\\text{products}} – \\sum H_{\\text{reactants}}
]<\/p>\n\n\n\nExample Reaction:<\/h3>\n\n\n\n
\\text{C} (s) + \\text{O}_2 (g) \\rightarrow \\text{CO}_2 (g)
]<\/p>\n\n\n\nSteps to Calculate ( \\Delta H ):<\/h3>\n\n\n\n
\n
\n
\n
[
\\Delta H = \\left( \\Delta H_f^\\circ \\, \\text{of products} \\right) – \\left( \\Delta H_f^\\circ \\, \\text{of reactants} \\right)
]
[
\\Delta H = \\left(-393.5 \\, \\text{kJ\/mol} \\right) – \\left(0 + 0\\right) = -393.5 \\, \\text{kJ\/mol}
]<\/li>\n<\/ol>\n\n\n\nConclusion:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n