{"id":256006,"date":"2025-07-13T12:47:07","date_gmt":"2025-07-13T12:47:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256006"},"modified":"2025-07-13T12:47:09","modified_gmt":"2025-07-13T12:47:09","slug":"use-the-data-provided-below-to-calculate-the-lattice-energy-of-rbcl","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/13\/use-the-data-provided-below-to-calculate-the-lattice-energy-of-rbcl\/","title":{"rendered":"Use the data provided below to calculate the lattice energy of RbCl."},"content":{"rendered":"\n

Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than the lattice energy of NaCl? Explain. Electron affinity of Cl = -349 kJ\/mol 1st ionization energy of Rb = 403 kJ\/mol Bond energy of Cl2 = 242 kJ\/mol Sublimation energy of Rb = 86.5 kJ\/mol \u00ce\u201dH\u00c2\u00b0[RbCl (s)] = -430.5 kJ\/mol Rb (s) + 1\/2 Cl2 (g) \u00e2\u2020\u2019 RbCl (s) This equation can be written as the sum of the following: Rb (s) \u00e2\u2020\u2019 Rb (g) sublimation \u00ce\u201dH\u00c2\u00b0 = +86.5 kJ\/mol Rb (g) \u00e2\u2020\u2019 Rb+ + e- 1st ionization energy \u00ce\u201dH\u00c2\u00b0 = +403 kJ\/mol Cl2 (g) \u00e2\u2020\u2019 2 Cl bond energy of Cl2 \u00ce\u201dH\u00c2\u00b0 = +242 kJ\/mol Cl + e- \u00e2\u2020\u2019 Cl- electron affinity of Cl \u00ce\u201dH\u00c2\u00b0 = -349 kJ\/mol Rb+ + Cl- \u00e2\u2020\u2019 RbCl lattice energy \u00ce\u201dH\u00c2\u00b0 = -222 kJ\/mol \u00ce\u201dH\u00c2\u00b0 = \u00ce\u201dH\u00c2\u00b0 + \u00ce\u201dH\u00c2\u00b0 + \u00ce\u201dH\u00c2\u00b0 + \u00ce\u201dH\u00c2\u00b0 + \u00ce\u201dH\u00c2\u00b0 Lattice energy \u00ce\u201dH\u00c2\u00b0 = \u00ce\u201dH\u00c2\u00b0 – (\u00ce\u201dH\u00c2\u00b0 + \u00ce\u201dH\u00c2\u00b0 + \u00ce\u201dH\u00c2\u00b0 + \u00ce\u201dH\u00c2\u00b0) = -430.5 – (86.5 + 403 + 242 – 349) = -692 kJ<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate the lattice energy of RbCl, we can break down the formation of RbCl from its elements using Hess’s Law, which states that the total enthalpy change for a process is the sum of the enthalpy changes for the steps into which the process can be divided.<\/p>\n\n\n\n

Step-by-Step Breakdown:<\/h3>\n\n\n\n

The formation of RbCl from its elements can be written as:<\/p>\n\n\n\n

Rb (s) + 1\/2 Cl2 (g) \u2192 RbCl (s)<\/strong><\/p>\n\n\n\n

This process can be broken into the following steps:<\/p>\n\n\n\n

    \n
  1. Sublimation of Rb (s) \u2192 Rb (g):<\/strong>
    The energy required to convert solid Rb into gaseous Rb is called the sublimation energy.
    \u0394H\u00b0 = +86.5 kJ\/mol<\/strong><\/li>\n\n\n\n
  2. Ionization of Rb (g) \u2192 Rb+ (g) + e-:<\/strong>
    The first ionization energy is the energy needed to remove an electron from a gaseous Rb atom to form Rb+.
    \u0394H\u00b0 = +403 kJ\/mol<\/strong><\/li>\n\n\n\n
  3. Dissociation of Cl2 (g) \u2192 2 Cl (g):<\/strong>
    The bond energy of Cl2 is the energy required to break the Cl-Cl bond and form two chlorine atoms in the gas phase.
    \u0394H\u00b0 = +242 kJ\/mol<\/strong><\/li>\n\n\n\n
  4. Electron affinity of Cl (g) + e- \u2192 Cl- (g):<\/strong>
    The electron affinity of chlorine is the energy released when a chlorine atom gains an electron to form Cl-.
    \u0394H\u00b0 = -349 kJ\/mol<\/strong><\/li>\n\n\n\n
  5. Formation of RbCl from Rb+ and Cl- (g):<\/strong>
    The lattice energy is the energy released when a solid ionic lattice forms from gaseous ions. This is the quantity we want to calculate.
    \u0394H\u00b0 = Lattice Energy<\/strong><\/li>\n<\/ol>\n\n\n\n

    Overall Enthalpy Change:<\/h3>\n\n\n\n

    The enthalpy change for the entire process of forming RbCl from its elements is given as:
    \u0394H\u00b0 = -430.5 kJ\/mol<\/strong><\/p>\n\n\n\n

    Now, applying Hess\u2019s Law, we can sum the enthalpies of all the steps:\u0394Hformation=\u0394Hsublimation+\u0394Hionization+\u0394Hbond energy+\u0394Helectron affinity+\u0394Hlattice\\Delta H_{\\text{formation}} = \\Delta H_{\\text{sublimation}} + \\Delta H_{\\text{ionization}} + \\Delta H_{\\text{bond energy}} + \\Delta H_{\\text{electron affinity}} + \\Delta H_{\\text{lattice}}\u0394Hformation\u200b=\u0394Hsublimation\u200b+\u0394Hionization\u200b+\u0394Hbond energy\u200b+\u0394Helectron affinity\u200b+\u0394Hlattice\u200b\u2212430.5=86.5+403+242\u2212349+\u0394Hlattice-430.5 = 86.5 + 403 + 242 – 349 + \\Delta H_{\\text{lattice}}\u2212430.5=86.5+403+242\u2212349+\u0394Hlattice\u200b<\/p>\n\n\n\n

    Simplifying the right side:\u2212430.5=86.5+403+242\u2212349+\u0394Hlattice-430.5 = 86.5 + 403 + 242 – 349 + \\Delta H_{\\text{lattice}}\u2212430.5=86.5+403+242\u2212349+\u0394Hlattice\u200b\u2212430.5=382.5+\u0394Hlattice-430.5 = 382.5 + \\Delta H_{\\text{lattice}}\u2212430.5=382.5+\u0394Hlattice\u200b\u0394Hlattice=\u2212430.5\u2212382.5=\u2212692 kJ\/mol\\Delta H_{\\text{lattice}} = -430.5 – 382.5 = -692 \\text{ kJ\/mol}\u0394Hlattice\u200b=\u2212430.5\u2212382.5=\u2212692 kJ\/mol<\/p>\n\n\n\n

    Comparison with NaCl:<\/h3>\n\n\n\n

    Lattice energy is influenced by two main factors: the charge<\/strong> of the ions and the size<\/strong> of the ions. The lattice energy increases with the charge of the ions and decreases with the size of the ions. Since Rb+ is larger than Na+ (because Rb is lower in the periodic table), the lattice energy of RbCl will be less<\/strong> than that of NaCl. Larger ions result in weaker electrostatic attraction between them, which means a smaller lattice energy.<\/p>\n\n\n\n

    Thus, the lattice energy of RbCl is -692 kJ\/mol<\/strong>, and it is less than<\/strong> the lattice energy of NaCl, which is typically around -786 kJ\/mol<\/strong>. This difference is due to the larger ionic radius of Rb+ compared to Na+, leading to a weaker ionic bond in RbCl.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than the lattice energy of NaCl? Explain. Electron affinity of Cl = -349 kJ\/mol 1st ionization energy of Rb = 403 kJ\/mol Bond energy of Cl2 = 242 kJ\/mol Sublimation energy of Rb = 86.5 kJ\/mol \u00ce\u201dH\u00c2\u00b0[RbCl […]<\/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-256006","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256006","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=256006"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256006\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=256006"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=256006"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=256006"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}