{"id":37614,"date":"2025-06-26T03:18:07","date_gmt":"2025-06-26T03:18:07","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37614"},"modified":"2025-06-26T03:18:08","modified_gmt":"2025-06-26T03:18:08","slug":"write-a-balanced-equation-for-the-combustion-of-c3h8g-propane","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/write-a-balanced-equation-for-the-combustion-of-c3h8g-propane\/","title":{"rendered":"Write a balanced equation for the combustion of C3H8(g) (propane)"},"content":{"rendered":"\n<p>Write a balanced equation for the combustion of C3H8(g) (propane) &#8211; i.e. its reaction with O2(g) forming the products CO2(g) and H2O(l). Given the following standard heats of formation: \u00ce\u201dHf\u00c2\u00b0 of CO2(g) is -393.5 kJ\/mol \u00ce\u201dHf\u00c2\u00b0 of H2O(l) is -286 kJ\/mol \u00ce\u201dHf\u00c2\u00b0 of C3H8(g) is -103.8 kJ\/mol What is the standard heat of reaction (\u00ce\u201dH\u00c2\u00b0) for the combustion reaction of C3H8(g)? Calculate the difference, \u00ce\u201dH-\u00ce\u201dE=\u00ce\u201d(PV) for the combustion reaction of 1 mole of propane. (Assume standard state conditions and 298 K for all reactants and products.)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Balanced Combustion Reaction of Propane (C\u2083H\u2088):<\/strong><\/h3>\n\n\n\n<p>The balanced chemical equation is:C3H8(g)+5O2(g)\u21923CO2(g)+4H2O(l)\\text{C}_3\\text{H}_8(g) + 5\\text{O}_2(g) \\rightarrow 3\\text{CO}_2(g) + 4\\text{H}_2\\text{O}(l)C3\u200bH8\u200b(g)+5O2\u200b(g)\u21923CO2\u200b(g)+4H2\u200bO(l)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Calculate the Standard Heat of Reaction (\u0394H\u00b0)<\/strong><\/h3>\n\n\n\n<p>Use the standard enthalpy of formation (\u0394Hf\u00b0) values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Hf\u00b0 [C\u2083H\u2088(g)] = -103.8 kJ\/mol<\/li>\n\n\n\n<li>\u0394Hf\u00b0 [CO\u2082(g)] = -393.5 kJ\/mol<\/li>\n\n\n\n<li>\u0394Hf\u00b0 [H\u2082O(l)] = -286 kJ\/mol<\/li>\n<\/ul>\n\n\n\n<p>Apply the formula:\u0394H\u2218=\u2211\u0394Hf\u2218(products)\u2212\u2211\u0394Hf\u2218(reactants)\\Delta H^\\circ = \\sum \\Delta H_f^\\circ \\text{(products)} &#8211; \\sum \\Delta H_f^\\circ \\text{(reactants)}\u0394H\u2218=\u2211\u0394Hf\u2218\u200b(products)\u2212\u2211\u0394Hf\u2218\u200b(reactants)=[3(\u2212393.5)+4(\u2212286)]\u2212[1(\u2212103.8)+5(0)]= \\left[3(-393.5) + 4(-286)\\right] &#8211; \\left[1(-103.8) + 5(0)\\right]=[3(\u2212393.5)+4(\u2212286)]\u2212[1(\u2212103.8)+5(0)]=[\u22121180.5\u22121144]\u2212[\u2212103.8]= [-1180.5 &#8211; 1144] &#8211; [-103.8]=[\u22121180.5\u22121144]\u2212[\u2212103.8]=\u22122324.5+103.8=\u22122220.7&nbsp;kJ\/mol= -2324.5 + 103.8 = -2220.7 \\text{ kJ\/mol}=\u22122324.5+103.8=\u22122220.7&nbsp;kJ\/mol<\/p>\n\n\n\n<p>So, the standard heat of combustion of propane is:\u0394H\u2218=\u22122220.7&nbsp;kJ\/mol\\boxed{\\Delta H^\\circ = -2220.7 \\text{ kJ\/mol}}\u0394H\u2218=\u22122220.7&nbsp;kJ\/mol\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Calculate \u0394H \u2013 \u0394E = \u0394(PV)<\/strong><\/h3>\n\n\n\n<p>Use the ideal gas law relationship:\u0394(PV)=\u0394ngas\u22c5RT\\Delta(PV) = \\Delta n_{\\text{gas}} \\cdot RT\u0394(PV)=\u0394ngas\u200b\u22c5RT<\/p>\n\n\n\n<p>At standard conditions (T = 298 K, R = 8.314 J\/mol\u00b7K = 0.008314 kJ\/mol\u00b7K):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Reactant gas moles: 1 C\u2083H\u2088 + 5 O\u2082 = 6 mol<\/li>\n\n\n\n<li>Product gas moles: 3 CO\u2082 (H\u2082O is liquid, not gas) = 3 mol<\/li>\n\n\n\n<li>\u0394n = n_products \u2013 n_reactants = 3 \u2013 6 = -3 mol<\/li>\n<\/ul>\n\n\n\n<p>Now calculate:\u0394(PV)=\u22123\u22c50.008314\u22c5298\u2248\u22127.43&nbsp;kJ\\Delta(PV) = -3 \\cdot 0.008314 \\cdot 298 \\approx -7.43 \\text{ kJ}\u0394(PV)=\u22123\u22c50.008314\u22c5298\u2248\u22127.43&nbsp;kJ<\/p>\n\n\n\n<p>So,\u0394H\u2212\u0394E=\u0394(PV)=\u22127.43&nbsp;kJ\\boxed{\\Delta H &#8211; \\Delta E = \\Delta(PV) = -7.43 \\text{ kJ}}\u0394H\u2212\u0394E=\u0394(PV)=\u22127.43&nbsp;kJ\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation)<\/strong><\/h3>\n\n\n\n<p>The combustion of propane (C\u2083H\u2088) is a chemical reaction in which propane reacts with oxygen (O\u2082) to form carbon dioxide (CO\u2082) and water (H\u2082O). This reaction is highly exothermic and releases a significant amount of energy in the form of heat.<\/p>\n\n\n\n<p>To begin, we write and balance the chemical equation to ensure conservation of mass. Each molecule of propane contains three carbon atoms and eight hydrogen atoms, so we need three CO\u2082 molecules and four H\u2082O molecules to balance carbon and hydrogen, respectively. Balancing oxygen then requires five O\u2082 molecules.<\/p>\n\n\n\n<p>To find the standard enthalpy change (\u0394H\u00b0), we use standard heats of formation. This is done by subtracting the total enthalpies of the reactants from those of the products. Since elements in their standard form, like O\u2082(g), have a \u0394Hf\u00b0 of zero, they do not contribute directly to the calculation. The calculation shows that burning one mole of propane releases 2220.7 kJ of heat.<\/p>\n\n\n\n<p>We also explore the difference between enthalpy (\u0394H) and internal energy (\u0394E) by computing the change in pressure-volume work (\u0394(PV)). This depends on the change in moles of gas. Since the number of gas molecules decreases (from 6 to 3), the system compresses, and the volume decreases, resulting in a negative \u0394(PV) value of -7.43 kJ. This small difference reflects the work done by the system due to volume change, which slightly reduces the energy released as internal energy.<\/p>\n\n\n\n<p>Overall, this analysis highlights how thermodynamic quantities relate to molecular changes in combustion and helps us understand energy conservation in chemical processes.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-932.jpeg\" alt=\"\" class=\"wp-image-37615\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-932.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-932-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-932-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Write a balanced equation for the combustion of C3H8(g) (propane) &#8211; i.e. its reaction with O2(g) forming the products CO2(g) and H2O(l). Given the following standard heats of formation: \u00ce\u201dHf\u00c2\u00b0 of CO2(g) is -393.5 kJ\/mol \u00ce\u201dHf\u00c2\u00b0 of H2O(l) is -286 kJ\/mol \u00ce\u201dHf\u00c2\u00b0 of C3H8(g) is -103.8 kJ\/mol What is the standard heat of reaction (\u00ce\u201dH\u00c2\u00b0) [&hellip;]<\/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-37614","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37614","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=37614"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37614\/revisions"}],"predecessor-version":[{"id":37616,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37614\/revisions\/37616"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37614"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37614"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37614"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}