Calculate Delta H r x n for the following reaction: Fe2O3 (s) + 3CO (g)—–> 2Fe (s) + 3CO2 (g). Use the following reactions and given delta H’s. 2Fe (s) + 3\/2O2 (g)—–>Fe2O3 (s), Delta H= -824.2 kJ CO (g) + 1\/2O2 (g)—–>CO2 (g), Delta H=-282.7 K j<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the enthalpy change ((\\Delta H_{rxn})) for the reaction:<\/p>\n\n\n\n [ using the given reactions:<\/p>\n\n\n\n The target reaction has Fe(_2)O(_3) as a reactant, but in the given equation, it appears as a product. So, we reverse the equation:<\/p>\n\n\n\n [ Since we reversed the reaction, we also reverse the sign of (\\Delta H):<\/p>\n\n\n\n [ The given second reaction forms CO(_2) from CO. Since the target equation involves 3 CO molecules, we multiply the entire equation by 3:<\/p>\n\n\n\n [ Since we multiplied the reaction by 3, we also multiply (\\Delta H) by 3:<\/p>\n\n\n\n [ Now, summing both reactions:<\/p>\n\n\n\n [ The (\\frac{3}{2}O_2 (g)) cancels from both sides, giving:<\/p>\n\n\n\n [ which matches the target reaction.<\/p>\n\n\n\n [ The enthalpy change for the reaction is (-23.9) kJ<\/strong>, meaning the reaction is exothermic<\/strong>, releasing a small amount of energy. This makes sense because iron oxide is being reduced while carbon monoxide is oxidized to carbon dioxide, a common thermodynamically favorable process in metallurgy.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate Delta H r x n for the following reaction: Fe2O3 (s) + 3CO (g)—–> 2Fe (s) + 3CO2 (g). Use the following reactions and given delta H’s. 2Fe (s) + 3\/2O2 (g)—–>Fe2O3 (s), Delta H= -824.2 kJ CO (g) + 1\/2O2 (g)—–>CO2 (g), Delta H=-282.7 K j The Correct Answer and Explanation is : […]<\/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-187379","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187379","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=187379"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187379\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187379"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187379"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187379"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{Fe}_2\\text{O}_3 (s) + 3CO (g) \\rightarrow 2Fe (s) + 3CO_2 (g)
]<\/p>\n\n\n\n\n
[
2Fe (s) + \\frac{3}{2}O_2 (g) \\rightarrow Fe_2O_3 (s) \\quad \\Delta H = -824.2 \\text{ kJ}
]<\/li>\n\n\n\n
[
CO (g) + \\frac{1}{2}O_2 (g) \\rightarrow CO_2 (g) \\quad \\Delta H = -282.7 \\text{ kJ}
]<\/li>\n<\/ol>\n\n\n\nStep 1: Reverse the First Equation<\/h3>\n\n\n\n
Fe_2O_3 (s) \\rightarrow 2Fe (s) + \\frac{3}{2}O_2 (g)
]<\/p>\n\n\n\n
\\Delta H = +824.2 \\text{ kJ}
]<\/p>\n\n\n\nStep 2: Multiply the Second Equation<\/h3>\n\n\n\n
3CO (g) + \\frac{3}{2}O_2 (g) \\rightarrow 3CO_2 (g)
]<\/p>\n\n\n\n
\\Delta H = 3 \\times (-282.7) = -848.1 \\text{ kJ}
]<\/p>\n\n\n\nStep 3: Add the Two Adjusted Equations<\/h3>\n\n\n\n
Fe_2O_3 (s) + \\frac{3}{2}O_2 (g) + 3CO (g) \\rightarrow 2Fe (s) + \\frac{3}{2}O_2 (g) + 3CO_2 (g)
]<\/p>\n\n\n\n
Fe_2O_3 (s) + 3CO (g) \\rightarrow 2Fe (s) + 3CO_2 (g)
]<\/p>\n\n\n\nStep 4: Sum the Enthalpy Changes<\/h3>\n\n\n\n
\\Delta H_{rxn} = 824.2 + (-848.1) = -23.9 \\text{ kJ}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n