Half-life equation for first-order reactions: t1\/2 = 0.693\/k where t1\/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s-1). What is the half-life of a first-order reaction with a rate constant of 1.00times10-4 s-1? Express your answer with the appropriate units. What is the rate constant of a first-order reaction that takes 538seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units. A certain first-order reaction has a rate constant of 1.00times10-3 s-1. How long will it take for the reactant concentration to drop to 1\/8 of its initial value? Express your answer with the appropriate units.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The half-life (( t_{1\/2} )) of a first-order reaction is constant and independent of the reactant concentration. It is calculated using the equation ( t_{1\/2} = 0.693\/k ), where ( k ) is the rate constant. For ( k = 1.00 \\times 10^{-4} \\, \\text{s}^{-1} ), the half-life is 6930 seconds, showing that the reaction takes this long for the reactant concentration to halve.<\/p>\n\n\n\n Conversely, when the half-life is known, the rate constant can be determined by rearranging the equation to ( k = 0.693\/t_{1\/2} ). For ( t_{1\/2} = 538 \\, \\text{s} ), ( k = 1.29 \\times 10^{-3} \\, \\text{s}^{-1} ), which quantifies the reaction speed.<\/p>\n\n\n\n For the final scenario, determining how long it takes for the concentration to drop to ( \\frac{1}{8} ) involves the integrated rate law for first-order reactions: ( \\ln([A]_0\/[A]) = k t ). A drop to ( \\frac{1}{8} ) implies ( [A]_0\/[A] = 8 ), and substituting ( k = 1.00 \\times 10^{-3} \\, \\text{s}^{-1} ) gives ( t = 2079 \\, \\text{s} ). This demonstrates that first-order reactions exhibit exponential decay, and the time required for a specific concentration change depends on ( k ) and the initial concentration ratio. These principles are central to understanding chemical kinetics in both experimental and practical applications.<\/p>\n","protected":false},"excerpt":{"rendered":" Half-life equation for first-order reactions: t1\/2 = 0.693\/k where t1\/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s-1). What is the half-life of a first-order reaction with a rate constant of 1.00times10-4 s-1? Express your answer with the appropriate units. What is the rate constant of a […]<\/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-184822","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184822","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=184822"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184822\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184822"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184822"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184822"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Calculations and Answers:<\/h3>\n\n\n\n
\n
[
t_{1\/2} = \\frac{0.693}{k}
]
Substituting ( k = 1.00 \\times 10^{-4} \\, \\text{s}^{-1} ):
[
t_{1\/2} = \\frac{0.693}{1.00 \\times 10^{-4}} = 6930 \\, \\text{s}
] Answer:<\/strong> ( t_{1\/2} = 6930 \\, \\text{s} ).<\/li>\n<\/ol>\n\n\n\n
\n\n\n\n\n
[
k = \\frac{0.693}{t_{1\/2}}
]
Substituting ( t_{1\/2} = 538 \\, \\text{s} ):
[
k = \\frac{0.693}{538} = 1.29 \\times 10^{-3} \\, \\text{s}^{-1}
] Answer:<\/strong> ( k = 1.29 \\times 10^{-3} \\, \\text{s}^{-1} ).<\/li>\n<\/ol>\n\n\n\n
\n\n\n\n\n
[
\\ln\\left(\\frac{[A]_0}{[A]}\\right) = k t
]
When the concentration drops to ( \\frac{1}{8} ), ( \\frac{[A]_0}{[A]} = 8 ). Substituting ( \\ln(8) = 2.079 ) and ( k = 1.00 \\times 10^{-3} \\, \\text{s}^{-1} ):
[
2.079 = (1.00 \\times 10^{-3}) t
]
Solve for ( t ):
[
t = \\frac{2.079}{1.00 \\times 10^{-3}} = 2079 \\, \\text{s}
] Answer:<\/strong> ( t = 2079 \\, \\text{s} ).<\/li>\n<\/ol>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n