{"id":190337,"date":"2025-02-12T06:36:35","date_gmt":"2025-02-12T06:36:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190337"},"modified":"2025-02-12T06:36:39","modified_gmt":"2025-02-12T06:36:39","slug":"for-each-event-indicate-how-the-concentration-of-each-species-in-the-chemical-equation-will-then-change-to-reach-equilibrium","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/12\/for-each-event-indicate-how-the-concentration-of-each-species-in-the-chemical-equation-will-then-change-to-reach-equilibrium\/","title":{"rendered":"For each event, indicate how the concentration of each species in the chemical equation will then change to reach equilibrium"},"content":{"rendered":"\n

For each event, indicate how the concentration of each species in the chemical equation will then change to reach equilibrium. 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.<\/p>\n\n\n\n

2CO(g)2CO(g)<\/p>\n\n\n\n

++<\/p>\n\n\n\n

O2(g)O2(g)<\/p>\n\n\n\n

?–??–?<\/p>\n\n\n\n

2CO2(g)2CO2(g)<\/p>\n\n\n\n

increasing the concentration of COCO<\/p>\n\n\n\n

increasing the concentration of CO2CO2<\/p>\n\n\n\n

decreasing the volume of the system<\/p>\n\n\n\n

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

For the reaction:<\/p>\n\n\n\n

[
2 \\text{CO} (g) + O_2 (g) \\rightleftharpoons 2 \\text{CO}_2 (g)
]<\/p>\n\n\n\n

We are given the following changes:<\/p>\n\n\n\n

    \n
  1. Increasing the concentration of CO (carbon monoxide):<\/strong>
    According to Le Chatelier\u2019s principle, if the concentration of a reactant is increased, the system will shift to the right to counteract this change by consuming more CO and O\u2082 to produce CO\u2082. Therefore, the concentration of CO will decrease, the concentration of O\u2082 will also decrease, and the concentration of CO\u2082 will increase. Changes:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • CO: Down arrow (\u2193)<\/li>\n\n\n\n
    • O\u2082: Down arrow (\u2193)<\/li>\n\n\n\n
    • CO\u2082: Up arrow (\u2191)<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Increasing the concentration of CO\u2082 (carbon dioxide):<\/strong>
        When the concentration of a product (CO\u2082) is increased, the system will shift to the left to counterbalance this change by consuming CO\u2082 and producing more CO and O\u2082. Hence, CO\u2082 concentration will decrease, and CO and O\u2082 concentrations will increase. Changes:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • CO: Up arrow (\u2191)<\/li>\n\n\n\n
        • O\u2082: Up arrow (\u2191)<\/li>\n\n\n\n
        • CO\u2082: Down arrow (\u2193)<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Decreasing the volume of the system:<\/strong>
            Decreasing the volume increases the pressure of the system. According to Le Chatelier\u2019s principle, the system will shift toward the side with fewer gas molecules to reduce the pressure. In this case, the left side (2 CO + 1 O\u2082) has 3 gas molecules, while the right side (2 CO\u2082) has only 2 gas molecules. Therefore, the system will shift to the right, increasing the concentration of CO\u2082 while decreasing the concentrations of CO and O\u2082. Changes:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • CO: Down arrow (\u2193)<\/li>\n\n\n\n
            • O\u2082: Down arrow (\u2193)<\/li>\n\n\n\n
            • CO\u2082: Up arrow (\u2191)<\/li>\n<\/ul>\n\n\n\n

              Summary of Changes:<\/h3>\n\n\n\n
                \n
              • Increasing CO concentration: CO \u2193, O\u2082 \u2193, CO\u2082 \u2191<\/li>\n\n\n\n
              • Increasing CO\u2082 concentration: CO \u2191, O\u2082 \u2191, CO\u2082 \u2193<\/li>\n\n\n\n
              • Decreasing volume: CO \u2193, O\u2082 \u2193, CO\u2082 \u2191<\/li>\n<\/ul>\n\n\n\n

                Let me now generate an image summarizing these changes.<\/p>\n\n\n\n

                Here is the diagram illustrating the changes in the chemical equilibrium for the given reactions. The shifts in concentration of CO, O\u2082, and CO\u2082 are represented based on the changes in the system. Let me know if you’d like any modifications or further clarification!<\/p>\n\n\n\n

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

                For each event, indicate how the concentration of each species in the chemical equation will then change to reach equilibrium. 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. 2CO(g)2CO(g) ++ O2(g)O2(g) ?–??–? 2CO2(g)2CO2(g) increasing the […]<\/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-190337","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190337","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=190337"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190337\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190337"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190337"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190337"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}