{"id":179278,"date":"2024-12-30T10:59:24","date_gmt":"2024-12-30T10:59:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=179278"},"modified":"2024-12-30T10:59:26","modified_gmt":"2024-12-30T10:59:26","slug":"from-your-knowledge-of-x-and-y-in-the-equation-as-well-as-the-rate-in-a-given-experiment-from-your-graph-calculate-k-from-your-data","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/12\/30\/from-your-knowledge-of-x-and-y-in-the-equation-as-well-as-the-rate-in-a-given-experiment-from-your-graph-calculate-k-from-your-data\/","title":{"rendered":"From your knowledge of X and Y in the equation (as well as the rate in a given experiment from your graph) calculate k from your data"},"content":{"rendered":"\n

From your knowledge of X and Y in the equation (as well as the rate in a given experiment from your graph) calculate k from your data. Rate= K[I^-]^x [S2O8^2-]^y
Using the pictures of the trials given, I need help answering this question.<\/p>\n\n\n\n

1.From your knowledge of X and Y in the equation (as well as the rate in a given experiment from your graph) calculate k from your data.
Rate= K[I^-]^x [S2O8^2-]^y<\/p>\n\n\n\n

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

3 4 5 6 7 Time (s) between appearances of color 147(5) 180 (5) 203 (5) 199 (5) Aliquot no. 1 2 3 4 5 6 7 Time (s) between appearances of color Cumulative times (s) 147 0 327 (5) 530 (5) 724 (5) 927 (5) 1134 (5) 1400 (5) Solution 1 4,7 x Solution 2,0XS -203 (52 -207 (5) 266 (5) Solution 4. Initial [S\u00e2\u201a\u201a0g2-] = 0.10 M; initial [I] = 0.025 M. Time experiment started g Cumulative times (s) Total moles of 2- S\u00e2\u201a\u201aO\u00c2\u00b2 consumed 2.0\u00c3\u2014104 4.0\u00c3\u2014104 6.0×104 3. What effect does changing the [S\u00e2\u201a\u201aO82] have on the reaction? Increases by about 2 8.0\u00c3\u201410-4 10.×10-4 12\u00c3\u201410-4 14\u00c3\u2014104 4. Write the rate law for this reaction that is consistent with your data. Rate = K[[ ] [\u00e2\u201a\ufffd\u00e2\u201a\u201a\u00e2\u201a\u201a] Calculations 1. Rate of reaction, A[S\u00e2\u201a\u201aO2-]\/At, as calculated from graphs (that is, from slopes of lines): -6 Solution 3.0 X105 Solution 4 Total moles of 2- S\u00e2\u201a\u201aO\u00c2\u00b2 consumed 2.0\u00c3\u201410 4 4.0\u00c3\u201410-4 6.0\u00c3\u201410-4 8.0\u00c3\u201410-4 10.×104 12\u00c3\u201410-4 14\u00c3\u201410-4 2. What effect does doubling the concentration of I have on the rate of this reaction? abat Jolle Decreases by about 2 skond 31 de onsdag<\/p>\n\n\n\n

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

To calculate the rate constant ( k ) in the rate law for the reaction, we need to use the given rate law and the data from the trials.<\/p>\n\n\n\n

Step-by-Step Process:<\/h3>\n\n\n\n
    \n
  1. Write the Rate Law:<\/strong>
    The general rate law is given by:
    [
    \\text{Rate} = k [I^-]^x [S_2O_8^{2-}]^y
    ]
    Here, ( x ) and ( y ) are the reaction orders with respect to ( [I^-] ) and ( [S_2O_8^{2-}] ), respectively. These can be determined experimentally, likely from previous steps.<\/li>\n\n\n\n
  2. Determine the Exponents ( x ) and ( y ):<\/strong>
    From the problem setup, it is likely that you are given or have determined the values of ( x ) and ( y ) (reaction orders). In the problem, you may need to use the changes in concentrations and corresponding changes in rates to determine ( x ) and ( y ). For example, doubling the concentration of ( I^- ) or ( S_2O_8^{2-} ) and observing how the rate changes helps you find these exponents.<\/li>\n\n\n\n
  3. Calculate the Rate Constant ( k ):<\/strong>
    Once the values of ( x ) and ( y ) are known, select one set of trial data where you know the initial concentrations of ( [I^-] ) and ( [S_2O_8^{2-}] ), as well as the corresponding rate. Use these values in the rate law formula to solve for ( k ). For example, using data from Trial 1:
    [
    \\text{Rate}_1 = k [I^-]^x [S_2O_8^{2-}]^y
    ]
    Rearranging to solve for ( k ):
    [
    k = \\frac{\\text{Rate}_1}{[I^-]^x [S_2O_8^{2-}]^y}
    ]
    Substitute the values from the trial data (rate, concentrations) to find ( k ).<\/li>\n\n\n\n
  4. Interpretation of Data:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Effect of Changing ( [S_2O_8^{2-}] ):<\/strong> Doubling the concentration of ( S_2O_8^{2-} ) increases the reaction rate, suggesting that the reaction order with respect to ( S_2O_8^{2-} ) (i.e., ( y )) is positive. From the graph, you can observe how the rate changes with increasing ( [S_2O_8^{2-}] ) and use this information to confirm ( y ).<\/li>\n\n\n\n
    • Effect of Changing ( [I^-] ):<\/strong> Doubling the concentration of ( I^- ) likely increases the reaction rate, so the order with respect to ( I^- ) (i.e., ( x )) is also positive. You can confirm this by examining how the rate changes with ( [I^-] ).<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Rate Law Consistency:<\/strong>
        Finally, verify that the calculated rate constant ( k ) is consistent across all trials. If it is, your rate law and exponents ( x ) and ( y ) are correct.<\/li>\n<\/ol>\n\n\n\n

        Final Answer:<\/h3>\n\n\n\n

        The rate constant ( k ) can be calculated from the rate law using the concentrations and the reaction rate from a chosen trial. Make sure to determine the correct reaction orders ( x ) and ( y ) by analyzing how the rate changes with changes in concentration. After finding ( k ), you can confirm it using data from multiple trials.<\/p>\n","protected":false},"excerpt":{"rendered":"

        From your knowledge of X and Y in the equation (as well as the rate in a given experiment from your graph) calculate k from your data. Rate= K[I^-]^x [S2O8^2-]^yUsing the pictures of the trials given, I need help answering this question. 1.From your knowledge of X and Y in the equation (as well as […]<\/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-179278","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/179278","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=179278"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/179278\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=179278"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=179278"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=179278"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}