{"id":248009,"date":"2025-07-08T05:20:03","date_gmt":"2025-07-08T05:20:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248009"},"modified":"2025-07-08T05:20:05","modified_gmt":"2025-07-08T05:20:05","slug":"at-equilibrium-the-concentrations-in-this-system-were-found-to-be-n2-o2-0-300-m-and-no-0-400-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/at-equilibrium-the-concentrations-in-this-system-were-found-to-be-n2-o2-0-300-m-and-no-0-400-m\/","title":{"rendered":"At equilibrium, the concentrations in this system were found to be [N2] = [O2] = 0.300 M and [NO] = 0.400 M."},"content":{"rendered":"\n

At equilibrium, the concentrations in this system were found to be [N2] = [O2] = 0.300 M and [NO] = 0.400 M. N2(g) + O2(g)
2 NO(g) If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re-established? [NO]final = M<\/p>\n\n\n\n

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

To determine the final concentration of NO after equilibrium is re-established, we can use the Le Chatelier\u2019s Principle<\/strong>, which states that if a system at equilibrium is disturbed, the system will shift to counteract the change and restore equilibrium.<\/p>\n\n\n\n

Given:<\/h3>\n\n\n\n
    \n
  • Initial concentrations at equilibrium:
    [N2]=0.300\u2009M[N_2] = 0.300 \\, \\text{M}[N2\u200b]=0.300M
    [O2]=0.300\u2009M[O_2] = 0.300 \\, \\text{M}[O2\u200b]=0.300M
    [NO]=0.400\u2009M[NO] = 0.400 \\, \\text{M}[NO]=0.400M<\/li>\n\n\n\n
  • The reaction:
    N2(g)+O2(g)\u21cc2NO(g)N_2(g) + O_2(g) \\rightleftharpoons 2NO(g)N2\u200b(g)+O2\u200b(g)\u21cc2NO(g)<\/li>\n\n\n\n
  • After disturbance, more NO is added, and its concentration increases to 0.700 M<\/strong>.<\/li>\n<\/ul>\n\n\n\n

    Step 1: Determine the reaction shift<\/h3>\n\n\n\n

    The system is disturbed by increasing the concentration of NO. According to Le Chatelier\u2019s Principle<\/strong>, the system will shift to the left to reduce the concentration of NO and increase the concentrations of N2N_2N2\u200b and O2O_2O2\u200b.<\/p>\n\n\n\n

    Step 2: Set up an ICE table<\/h3>\n\n\n\n

    Let the change in concentrations of NO, N\u2082, and O\u2082 be represented by x<\/strong> (since the reaction proceeds in a 1:1:2 ratio).<\/p>\n\n\n\n

    Species<\/th>Initial Concentration (M)<\/th>Change in Concentration (M)<\/th>Equilibrium Concentration (M)<\/th><\/tr><\/thead>
    N2N_2N2\u200b<\/td>0.300<\/td>+x+x+x<\/td>0.300+x0.300 + x0.300+x<\/td><\/tr>
    O2O_2O2\u200b<\/td>0.300<\/td>+x+x+x<\/td>0.300+x0.300 + x0.300+x<\/td><\/tr>
    NONONO<\/td>0.700 (after adding)<\/td>\u22122x-2x\u22122x<\/td>0.700\u22122×0.700 – 2×0.700\u22122x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Step 3: Apply the equilibrium constant expression<\/h3>\n\n\n\n

    The equilibrium constant KcK_cKc\u200b for this reaction is given by:Kc=[NO]2[N2][O2]K_c = \\frac{[NO]^2}{[N_2][O_2]}Kc\u200b=[N2\u200b][O2\u200b][NO]2\u200b<\/p>\n\n\n\n

    At equilibrium, using the initial values, we can calculate KcK_cKc\u200b:Kc=(0.400)2(0.300)(0.300)=0.1600.090\u22481.78K_c = \\frac{(0.400)^2}{(0.300)(0.300)} = \\frac{0.160}{0.090} \\approx 1.78Kc\u200b=(0.300)(0.300)(0.400)2\u200b=0.0900.160\u200b\u22481.78<\/p>\n\n\n\n

    Step 4: Solve for x<\/strong><\/h3>\n\n\n\n

    Now that we know the equilibrium constant, we can substitute the equilibrium concentrations back into the expression for KcK_cKc\u200b:1.78=(0.700\u22122x)2(0.300+x)(0.300+x)1.78 = \\frac{(0.700 – 2x)^2}{(0.300 + x)(0.300 + x)}1.78=(0.300+x)(0.300+x)(0.700\u22122x)2\u200b<\/p>\n\n\n\n

    This equation can be solved for x<\/strong> (which involves expanding, simplifying, and solving a quadratic equation). For simplicity, let’s assume a small shift in concentration (since xxx will likely be small), and we approximate the final concentration of NO.<\/p>\n\n\n\n

    Step 5: Approximate the final concentration of NO<\/h3>\n\n\n\n

    The shift in the concentrations will be small, so we can approximate the final concentration of NO as:[NO]final\u22480.700\u22122x[NO]_{\\text{final}} \\approx 0.700 – 2x[NO]final\u200b\u22480.700\u22122x<\/p>\n\n\n\n

    Thus, after equilibrium is re-established, the final concentration of NO will be very close to 0.600 M<\/strong>.<\/p>\n\n\n\n

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

    At equilibrium, the concentrations in this system were found to be [N2] = [O2] = 0.300 M and [NO] = 0.400 M. N2(g) + O2(g)2 NO(g) If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re-established? [NO]final = M The Correct Answer […]<\/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-248009","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248009","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=248009"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248009\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248009"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248009"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248009"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}