Solution:<\/h3>\n\n\n\n
To determine the percent error<\/strong> in the experiment, we need to follow these steps:<\/p>\n\n\n\n Repeat similar calculations for Trials 2 and 3.<\/p>\n\n\n\n After determining \u0394H for each trial, you calculate the average \u0394H<\/strong> in kJ\/mol.<\/p>\n\n\n\n If the average measured \u0394H is, for example, -56.2 kJ\/mol, then: Percent Error=(\u2223\u221256.2\u2212(\u221257.0)\u222357.0)\u00d7100=(0.857.0)\u00d7100=1.4%\\text{Percent Error} = \\left( \\frac{\\left| -56.2 – (-57.0) \\right|}{57.0} \\right) \\times 100 = \\left( \\frac{0.8}{57.0} \\right) \\times 100 = 1.4\\%Percent Error=(57.0\u2223\u221256.2\u2212(\u221257.0)\u2223\u200b)\u00d7100=(57.00.8\u200b)\u00d7100=1.4%<\/p>\n\n\n\n After calculating the average \u0394H for your trials and comparing it to the theoretical value (-57.0 kJ\/mol), you can determine the percent error using the formula provided. In the example above, the percent error was 1.4%.<\/p>\n\n\n\n Many neutralization reactions are exothermic. In part 3, we will determine the heat of neutralization of hydrochloric acid with sodium hydroxide according to the reaction: HCl (aq) + NaOH (aq) \u00e2\u2020\u2019 NaCl (aq) + H2O (l) This reaction takes place in the same calorimeter as above, and we can summarize the heat flow the same […]<\/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-263268","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263268","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=263268"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263268\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=263268"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=263268"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=263268"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
The formula for heat flow is: qrxn=qwater+qcalorimeterq_{\\text{rxn}} = q_{\\text{water}} + q_{\\text{calorimeter}}qrxn\u200b=qwater\u200b+qcalorimeter\u200b Since both the heat absorbed by the water and the calorimeter are proportional to the change in temperature, the calculation for enthalpy per trial will be: q=m\u22c5C\u22c5\u0394Tq = m \\cdot C \\cdot \\Delta Tq=m\u22c5C\u22c5\u0394T where:\n\n
To convert \u0394H (in J\/g) to kJ\/mol, you need to account for the molar mass of the substance involved. The molarity of NaOH and HCl is 3 M, so we calculate moles and then determine the enthalpy in kJ\/mol. moles\u00a0of\u00a0NaOH=volume\u00a0of\u00a0NaOH\u00d7molarity\\text{moles of NaOH} = \\text{volume of NaOH} \\times \\text{molarity}moles\u00a0of\u00a0NaOH=volume\u00a0of\u00a0NaOH\u00d7molarity \u0394H\u00a0in\u00a0kJ\/mol=\u0394H\u00a0in\u00a0J\/g\u00d7molar\u00a0mass\u00a0of\u00a0NaOHmoles\u00a0of\u00a0NaOH\u00d710\u22123\\text{\u0394H in kJ\/mol} = \\frac{\\text{\u0394H in J\/g} \\times \\text{molar mass of NaOH}}{\\text{moles of NaOH}} \\times 10^{-3}\u0394H\u00a0in\u00a0kJ\/mol=moles\u00a0of\u00a0NaOH\u0394H\u00a0in\u00a0J\/g\u00d7molar\u00a0mass\u00a0of\u00a0NaOH\u200b\u00d710\u22123<\/li>\n\n\n\n
After calculating the enthalpy change for each trial, take the average value of the enthalpy changes in kJ\/mol.<\/li>\n\n\n\n
The percent error<\/strong> can be calculated as follows: Percent\u00a0Error=(\u2223Measured\u00a0\u0394H\u2212Theoretical\u00a0\u0394H\u2223Theoretical\u00a0\u0394H)\u00d7100\\text{Percent Error} = \\left( \\frac{\\left|\\text{Measured \u0394H} – \\text{Theoretical \u0394H}\\right|}{\\text{Theoretical \u0394H}} \\right) \\times 100Percent\u00a0Error=(Theoretical\u00a0\u0394H\u2223Measured\u00a0\u0394H\u2212Theoretical\u00a0\u0394H\u2223\u200b)\u00d7100 Where:\n\n
Now, let’s walk through the calculations:<\/h3>\n\n\n\n
Step 1: Calculate \u0394H in J\/g for each trial<\/h4>\n\n\n\n
\n
\u0394T=34.0\u00b0C\u221221.5\u00b0C=12.5\u00b0C\\Delta T = 34.0\u00b0C – 21.5\u00b0C = 12.5\u00b0C\u0394T=34.0\u00b0C\u221221.5\u00b0C=12.5\u00b0C
Using the given values for the mass (approximately 50 mL solution = 50 g) and the specific heat of water (4.18 J\/g\u00b7\u00b0C), we calculate: qrxn=50\u22c54.18\u22c512.5=2605\u00a0Jq_{\\text{rxn}} = 50 \\cdot 4.18 \\cdot 12.5 = 2605 \\text{ J}qrxn\u200b=50\u22c54.18\u22c512.5=2605\u00a0J Then calculate \u0394H in J\/g by dividing by the mass of the solution: \u0394Htrial\u00a01=260550=52.1\u00a0J\/g\\Delta H_{\\text{trial 1}} = \\frac{2605}{50} = 52.1 \\text{ J\/g}\u0394Htrial\u00a01\u200b=502605\u200b=52.1\u00a0J\/g<\/li>\n<\/ul>\n\n\n\nStep 2: Calculate the Average \u0394H in kJ\/mol<\/h4>\n\n\n\n
Step 3: Determine Percent Error<\/h4>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1291.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"