{"id":201335,"date":"2025-03-15T08:39:23","date_gmt":"2025-03-15T08:39:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201335"},"modified":"2025-03-15T08:39:25","modified_gmt":"2025-03-15T08:39:25","slug":"calculate-the-standard-enthalpy-of-reaction-for-the-reaction-2na-2h2o-2naoh-h2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/15\/calculate-the-standard-enthalpy-of-reaction-for-the-reaction-2na-2h2o-2naoh-h2\/","title":{"rendered":"Calculate the standard enthalpy of reaction for the reaction 2Na + 2H2O\u2014\u2014> 2NaOH+ H2"},"content":{"rendered":"\n<p>Calculate the standard enthalpy of reaction for the reaction 2Na + 2H2O\u2014\u2014&gt; 2NaOH+ H2. Standard enthalpies of formation are -285.8 kJ\/mol for H2O and -470.11 kJ\/mol for NaOH.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To calculate the standard enthalpy of reaction (\u0394Hreaction\u2218\\Delta H^\\circ_{\\text{reaction}}) for the given reaction: 2Na+2H2O\u21922NaOH+H22 \\text{Na} + 2 \\text{H}_2\\text{O} \\rightarrow 2 \\text{NaOH} + \\text{H}_2<\/p>\n\n\n\n<p>we will use the standard enthalpies of formation (\u0394Hf\u2218\\Delta H^\\circ_f) of the reactants and products involved. The formula to calculate the standard enthalpy of the reaction is: \u0394Hreaction\u2218=\u2211\u0394Hf\u2218(products)\u2212\u2211\u0394Hf\u2218(reactants)\\Delta H^\\circ_{\\text{reaction}} = \\sum \\Delta H^\\circ_f (\\text{products}) &#8211; \\sum \\Delta H^\\circ_f (\\text{reactants})<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: List the standard enthalpies of formation for each compound involved.<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Standard enthalpy of formation of Na\\text{Na} (sodium) is 0\u2009kJ\/mol0 \\, \\text{kJ\/mol} (since it is in its elemental form).<\/li>\n\n\n\n<li>Standard enthalpy of formation of H2O\\text{H}_2\\text{O} (liquid water) is \u2212285.8\u2009kJ\/mol-285.8 \\, \\text{kJ\/mol}.<\/li>\n\n\n\n<li>Standard enthalpy of formation of NaOH\\text{NaOH} (sodium hydroxide) is \u2212470.11\u2009kJ\/mol-470.11 \\, \\text{kJ\/mol}.<\/li>\n\n\n\n<li>Standard enthalpy of formation of H2\\text{H}_2 (hydrogen gas) is 0\u2009kJ\/mol0 \\, \\text{kJ\/mol} (since it is in its elemental form).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the sum of the enthalpies of formation for the products and reactants.<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Products:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For 2NaOH2 \\text{NaOH}, the total enthalpy is:<\/li>\n<\/ul>\n\n\n\n<p>2\u00d7(\u2212470.11\u2009kJ\/mol)=\u2212940.22\u2009kJ2 \\times (-470.11 \\, \\text{kJ\/mol}) = -940.22 \\, \\text{kJ}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For H2\\text{H}_2, the total enthalpy is:<\/li>\n<\/ul>\n\n\n\n<p>1\u00d70\u2009kJ\/mol=0\u2009kJ1 \\times 0 \\, \\text{kJ\/mol} = 0 \\, \\text{kJ}<\/p>\n\n\n\n<p>Thus, the total enthalpy for the products is: \u2212940.22\u2009kJ-940.22 \\, \\text{kJ}<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Reactants:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For 2Na2 \\text{Na}, the total enthalpy is:<\/li>\n<\/ul>\n\n\n\n<p>2\u00d70\u2009kJ\/mol=0\u2009kJ2 \\times 0 \\, \\text{kJ\/mol} = 0 \\, \\text{kJ}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For 2H2O2 \\text{H}_2\\text{O}, the total enthalpy is:<\/li>\n<\/ul>\n\n\n\n<p>2\u00d7(\u2212285.8\u2009kJ\/mol)=\u2212571.6\u2009kJ2 \\times (-285.8 \\, \\text{kJ\/mol}) = -571.6 \\, \\text{kJ}<\/p>\n\n\n\n<p>Thus, the total enthalpy for the reactants is: 0\u2009kJ+(\u2212571.6\u2009kJ)=\u2212571.6\u2009kJ0 \\, \\text{kJ} + (-571.6 \\, \\text{kJ}) = -571.6 \\, \\text{kJ}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the standard enthalpy of the reaction.<\/h3>\n\n\n\n<p>Now, apply the formula for the enthalpy of reaction: \u0394Hreaction\u2218=[\u2212940.22\u2009kJ]\u2212[\u2212571.6\u2009kJ]\\Delta H^\\circ_{\\text{reaction}} = \\left[ -940.22 \\, \\text{kJ} \\right] &#8211; \\left[ -571.6 \\, \\text{kJ} \\right] \u0394Hreaction\u2218=\u2212940.22\u2009kJ+571.6\u2009kJ\\Delta H^\\circ_{\\text{reaction}} = -940.22 \\, \\text{kJ} + 571.6 \\, \\text{kJ} \u0394Hreaction\u2218=\u2212368.62\u2009kJ\\Delta H^\\circ_{\\text{reaction}} = -368.62 \\, \\text{kJ}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>The standard enthalpy of reaction for the given reaction is \u0394Hreaction\u2218=\u2212368.62\u2009kJ\\Delta H^\\circ_{\\text{reaction}} = -368.62 \\, \\text{kJ}.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The standard enthalpy of reaction reflects the heat absorbed or released when the reaction occurs under standard conditions (298 K and 1 atm). In this reaction, sodium reacts with water to form sodium hydroxide and hydrogen gas. Since the enthalpy of formation for elements in their natural states (Na and H2\\text{H}_2) is zero, the overall enthalpy change depends primarily on the enthalpies of formation of NaOH\\text{NaOH} and H2O\\text{H}_2\\text{O}. The negative sign indicates that the reaction is exothermic, meaning it releases energy to the surroundings.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the standard enthalpy of reaction for the reaction 2Na + 2H2O\u2014\u2014&gt; 2NaOH+ H2. Standard enthalpies of formation are -285.8 kJ\/mol for H2O and -470.11 kJ\/mol for NaOH. The correct answer and explanation is : To calculate the standard enthalpy of reaction (\u0394Hreaction\u2218\\Delta H^\\circ_{\\text{reaction}}) for the given reaction: 2Na+2H2O\u21922NaOH+H22 \\text{Na} + 2 \\text{H}_2\\text{O} \\rightarrow 2 [&hellip;]<\/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-201335","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201335","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=201335"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201335\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201335"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201335"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201335"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}