{"id":148933,"date":"2024-02-27T18:09:40","date_gmt":"2024-02-27T12:39:40","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=148933"},"modified":"2024-02-27T18:09:40","modified_gmt":"2024-02-27T12:39:40","slug":"show-that-the-time-required-for-99-9-completion-in-a-first-order-reaction-is-10-times-of-half-lif","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/quiz\/show-that-the-time-required-for-99-9-completion-in-a-first-order-reaction-is-10-times-of-half-lif","title":{"rendered":"Show that the time required for 99.9% completion in a first-order reaction is 10 times of half-lif…"},"content":{"rendered":"

Show that the time required for 99.9% completion in a first-order reaction is 10 times of half-life (t 1\/2) of the reaction [log 2 = 0.3010, log 10 = 1].<\/h3>\n

Ans.<\/h3>\n
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Let’s summarize the given information and derivation:<\/h3>\n

For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The integrated rate law for a first-order reaction is:<\/h3>\n

ln\u2061([\ufffd][\ufffd]0)=\u2212\ufffd\ufffd<\/span>ln<\/span>(<\/span><\/span>[<\/span>A<\/span>]0<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>[<\/span>A<\/span>]<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span>=<\/span><\/span>\u2212<\/span>k<\/span>t<\/span><\/span><\/span><\/span><\/span><\/h3>\n

Where:<\/h3>\n
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    [\ufffd]<\/span>[<\/span>A<\/span>]<\/span><\/span><\/span><\/span><\/span> is the concentration of the reactant at time \ufffd<\/span>t<\/span><\/span><\/span><\/span><\/span>,<\/h3>\n<\/li>\n
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    [\ufffd]0<\/span>[<\/span>A<\/span>]0<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the initial concentration of the reactant,<\/h3>\n<\/li>\n
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    \ufffd<\/span>k<\/span><\/span><\/span><\/span><\/span> is the rate constant of the reaction,<\/h3>\n<\/li>\n
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    \ufffd<\/span>t<\/span><\/span><\/span><\/span><\/span> is the time.<\/h3>\n<\/li>\n<\/ul>\n

    The half-life (\ufffd1\/2<\/span>t<\/span>1\/2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>) of a first-order reaction is given by:<\/h3>\n

    \ufffd1\/2=ln\u20612\ufffd<\/span>t<\/span>1\/2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>k<\/span><\/span><\/span>l<\/span>n<\/span><\/span>2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

    Given that log\u20612=0.3010<\/span>log<\/span>2<\/span>=<\/span><\/span>0.3010<\/span><\/span><\/span><\/span><\/span>, we can rewrite the equation for the half-life as:<\/h3>\n

    \ufffd1\/2=0.3010\ufffd<\/span>t<\/span>1\/2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>k<\/span><\/span><\/span>0.3010<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

    Now, if we want to find the time required for 99.9% completion of the reaction, which means that only 0.1% of the reactant remains, we can express this as:<\/h3>\n

    ln\u2061([\ufffd][\ufffd]0)=\u2212ln\u2061(0.001)<\/span>ln<\/span>(<\/span><\/span>[<\/span>A<\/span>]0<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>[<\/span>A<\/span>]<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span>=<\/span><\/span>\u2212<\/span>ln<\/span>(<\/span>0.001<\/span>)<\/span><\/span><\/span><\/span><\/span><\/h3>\n

    Using the integrated rate law for a first-order reaction, we can set ln\u2061([\ufffd][\ufffd]0)=\u2212\ufffd\ufffd<\/span>ln<\/span>(<\/span><\/span>[<\/span>A<\/span>]0<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>[<\/span>A<\/span>]<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span>=<\/span><\/span>\u2212<\/span>k<\/span>t<\/span><\/span><\/span><\/span><\/span> and solve for \ufffd<\/span>t<\/span><\/span><\/span><\/span><\/span>:<\/h3>\n

    \u2212\ufffd\ufffd=\u2212ln\u2061(0.001)<\/span>\u2212<\/span>k<\/span>t<\/span>=<\/span><\/span>\u2212<\/span>ln<\/span>(<\/span>0.001<\/span>)<\/span><\/span><\/span><\/span><\/span> \ufffd\ufffd=ln\u2061(1000)<\/span>k<\/span>t<\/span>=<\/span><\/span>ln<\/span>(<\/span>1000<\/span>)<\/span><\/span><\/span><\/span><\/span> \ufffd=ln\u2061(1000)\ufffd<\/span>t<\/span>=<\/span><\/span>k<\/span><\/span><\/span>l<\/span>n<\/span><\/span>(<\/span>1000)<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

    We know that ln\u2061(1000)=3ln\u2061(10)=3<\/span>ln<\/span>(<\/span>1000<\/span>)<\/span>=<\/span><\/span>3<\/span>ln<\/span>(<\/span>10<\/span>)<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span><\/span>, since ln\u2061(10)=1<\/span>ln<\/span>(<\/span>10<\/span>)<\/span>=<\/span><\/span>1<\/span><\/span><\/span><\/span><\/span> (given that log\u206110=1<\/span>log<\/span>10<\/span>=<\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>).<\/h3>\n

    Therefore, we have: \ufffd=3\ufffd<\/span>t<\/span>=<\/span><\/span>k<\/span><\/span><\/span>3<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

    We also know that \ufffd1\/2=0.3010\ufffd<\/span>t<\/span>1\/2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>k<\/span><\/span><\/span>0.3010<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/h3>\n

    So, the time required for 99.9% completion is: \ufffd=10\u00d7\ufffd1\/2<\/span>t<\/span>=<\/span><\/span>10<\/span>\u00d7<\/span><\/span>t<\/span>1\/2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

    Thus, we have demonstrated that the time required for 99.9% completion in a first-order reaction is 10 times the half-life (\ufffd1\/2<\/span>t<\/span>1\/2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>) of the reaction.<\/h3>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    Show that the time required for 99.9% completion in a first-order reaction is 10 times of half-life (t 1\/2) of the reaction [log 2 = 0.3010, log 10 = 1]. Ans. Let’s summarize the given information and derivation: For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. […]<\/p>\n","protected":false},"author":21830,"featured_media":114041,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12119],"tags":[],"class_list":{"0":"post-148933","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-quiz"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148933","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=148933"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148933\/revisions"}],"predecessor-version":[{"id":148934,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148933\/revisions\/148934"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/114041"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=148933"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=148933"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=148933"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}