{"id":239295,"date":"2025-07-02T19:35:11","date_gmt":"2025-07-02T19:35:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239295"},"modified":"2025-07-02T19:35:13","modified_gmt":"2025-07-02T19:35:13","slug":"indicate-how-the-concentration-of-each-species-in-the-chemical-equation-will-change-to-reestablish-equilibrium-after-reactant-or-product-is-added-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/02\/indicate-how-the-concentration-of-each-species-in-the-chemical-equation-will-change-to-reestablish-equilibrium-after-reactant-or-product-is-added-2\/","title":{"rendered":"Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added"},"content":{"rendered":"\n

Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added. An up arrow indicates an increase in concentration, a down arrow indicates a decrease in concentration, and leaving it blank means there is no change in the concentration.
increasing the concentration of CO increasing the concentration of CO
Answer Bank<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

Here are the correct answers indicating the change in concentration for each species, followed by a detailed explanation.<\/p>\n\n\n\n

The chemical equation is: 2CO(g) + O\u2082(g) \u21cc 2CO\u2082(g)<\/strong><\/p>\n\n\n\n

Condition<\/td>Change in [CO]<\/td>Change in [O\u2082]<\/td>Change in [CO\u2082]<\/td><\/tr>
increasing the concentration of CO<\/strong><\/td>\u2191<\/td>\u2193<\/td>\u2191<\/td><\/tr>
increasing the concentration of CO\u2082<\/strong><\/td>\u2191<\/td>\u2191<\/td>\u2191<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Explanation<\/h3>\n\n\n\n

This problem is solved using Le Ch\u00e2telier’s principle, which states that if a change (or stress) is applied to a system at equilibrium, the system will shift in a direction that partially counteracts the stress to establish a new equilibrium.<\/p>\n\n\n\n

1. Increasing the concentration of CO<\/strong><\/p>\n\n\n\n

    \n
  • Stress:<\/strong>\u00a0The concentration of a reactant, carbon monoxide (CO), is increased.<\/li>\n\n\n\n
  • System’s Response:<\/strong>\u00a0To counteract this stress, the system will shift to consume the added CO. This means the forward reaction is favored. The equilibrium shifts to the\u00a0right<\/strong>.<\/li>\n\n\n\n
  • Concentration Changes:<\/strong>\n
      \n
    • CO:<\/strong>\u00a0The system is stressed by adding CO. Although the forward reaction consumes some CO to reach the new equilibrium, it does not consume all of the amount that was added. Therefore, the final concentration of CO at the new equilibrium will be\u00a0higher<\/strong>\u00a0(\u2191) than its original concentration.<\/li>\n\n\n\n
    • O\u2082:<\/strong>\u00a0As the equilibrium shifts to the right, the other reactant, oxygen (O\u2082), is consumed. Its concentration will\u00a0decrease<\/strong>\u00a0(\u2193).<\/li>\n\n\n\n
    • CO\u2082:<\/strong>\u00a0The shift to the right favors the formation of the product, carbon dioxide (CO\u2082). Its concentration will\u00a0increase<\/strong>\u00a0(\u2191).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

      2. Increasing the concentration of CO\u2082<\/strong><\/p>\n\n\n\n

        \n
      • Stress:<\/strong>\u00a0The concentration of the product, carbon dioxide (CO\u2082), is increased.<\/li>\n\n\n\n
      • System’s Response:<\/strong>\u00a0To relieve this stress, the system will shift to consume the added CO\u2082. This means the reverse reaction is favored. The equilibrium shifts to the\u00a0left<\/strong>.<\/li>\n\n\n\n
      • Concentration Changes:<\/strong>\n
          \n
        • CO\u2082:<\/strong>\u00a0Similar to the first scenario, the system is stressed by adding CO\u2082. The reverse reaction consumes some of the added CO\u2082, but the new equilibrium concentration will still be\u00a0higher<\/strong>\u00a0(\u2191) than the original concentration.<\/li>\n\n\n\n
        • CO:<\/strong>\u00a0As the equilibrium shifts to the left, the reactant carbon monoxide (CO) is produced. Its concentration will\u00a0increase<\/strong>\u00a0(\u2191).<\/li>\n\n\n\n
        • O\u2082:<\/strong>\u00a0The shift to the left also produces the reactant oxygen (O\u2082). Its concentration will also\u00a0increase<\/strong>\u00a0(\u2191).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added. An up arrow indicates an increase in concentration, a down arrow indicates a decrease in concentration, and leaving it blank means there is no change in the concentration.increasing the concentration of CO increasing […]<\/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-239295","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239295","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=239295"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239295\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239295"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239295"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239295"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}