{"id":160,"date":"2017-06-20T18:24:21","date_gmt":"2017-06-20T18:24:21","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/?post_type=chapter&p=160"},"modified":"2017-06-20T21:14:01","modified_gmt":"2017-06-20T21:14:01","slug":"lab-6-introduction","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/chapter\/lab-6-introduction\/","title":{"raw":"Lab 6 Introduction","rendered":"Lab 6 Introduction"},"content":{"raw":"
\r\n
\r\n
\r\n\r\nChesapeake Campus \u2013 Chemistry 112 Laboratory\r\n\r\nLab #6- Kinetics: Reaction Rates and Steady State Approximation\r\n\r\nObjectives\r\n
    \r\n \t
  • \uf0b7 \u00a0Determine the rate law for a chemical equation.<\/li>\r\n \t
  • \uf0b7 \u00a0Evaluate a two-step reaction with different rate constants using dice.<\/li>\r\n \t
  • \uf0b7 \u00a0Determine the rate-determining step of a reaction.<\/li>\r\n \t
  • \uf0b7 \u00a0Calculate the rate constant for a reaction.IntroductionKineticsKinetics is the study of how fast a chemical reaction occurs. Some reactions occur very quickly. Consider mixing an aqueous solution of silver nitrate and sodium chloride. As soon as these chemicals touch there is an immediate appearance of the precipitated silver chloride. On the other hand some reactions occur very slowly (even if they are thermodynamically preferred). Technically the decay of diamond to graphite is thermodynamically favored but the reaction has a large activation energy. Due to the activation energy, the reaction takes place extremely slowly leading to the belief that a diamond lasts forever.Rate Law\r\n\r\nA rate law is a mathematical description of how fast a chemical reaction takes place. For a given chemical equation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n\r\nthe given rate law is defined as\r\n\r\n<\/div>\r\n
    \r\n\r\naA + bB \u2192 cC\r\n\r\n\ud835\udc5f\ud835\udc4e\ud835\udc61\ud835\udc52 = \ud835\udc58[\ud835\udc34]\ud835\udc65[\ud835\udc35]\ud835\udc66\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n\r\nwhere the rate is defined as being equal to the rate constant (k) multiplied by the concentrations of all reactants raised to the exponents that are different for each chemical reaction. The larger the value of k, generally the faster the reaction occurs. The order of the reaction is equal to the sum of all the exponents. For this experiment you will be working with first order reactions. You will experimentally determine the rate constant for a two distinct situations.\r\n\r\nThere are often reactions that occur in several steps. Consider the situation where a reactant decays into a series of products. We can illustrate this situation with an equation:\r\n\r\n\ud835\udc581 \ud835\udc582 \ud835\udc4e\ud835\udc34\u2192 \ud835\udc4f\ud835\udc35\u2192 \ud835\udc50\ud835\udc36\r\n\r\nHere we have a reactant A which decays or reacts to form B. This reaction has the rate constant k1 and can be expressed in terms of the change in concentration of A over time:\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n\r\n\ud835\udc51[\ud835\udc34] = \u2212\ud835\udc581[\ud835\udc34] \ud835\udc51\ud835\udc61\r\n\r\nHere the change in the concentration of A over time is equal to the negative rate constant multiplied by the concentration of A at the given point in time, t. The negative indicates the value of A is being lost as time progresses. The integrated rate law could be written as\r\n\r\n[\ud835\udc34] = [\ud835\udc34]0 \ud835\udc52\u2212\ud835\udc58\ud835\udc61 \ud835\udc5c\ud835\udc5f ln[\ud835\udc34] = ln[\ud835\udc34]0 \u2212 \ud835\udc58\ud835\udc61\r\n\r\nIn a similar way we can evaluate the second step of the reaction indicated by k2. The best way to do this would be to evaluate the amount of C produced as a function of the rate constant multiplied by the amount of reactant (B) available to react in a given time (t).\r\n\r\n\ud835\udc51[\ud835\udc36] = \ud835\udc582[\ud835\udc35] \ud835\udc51\ud835\udc61\r\n\r\nNote that there is no negative which will allow the value of C to go up. To determine the amount of B produced in this chemical process we would have to evaluate BOTH how much B is produced from the first step as well as the amount of B lost in the second step.\r\n\r\nRate-Determining Steps\r\n\r\nConsider the process above. If k2 >>> k1, then as soon as B was created, it would instantly react to make C. The rate determining step of the entire process would be k1 because it would be the slower of the two processes. In addition, because B is used up as quickly as it is made, we can assume the differential-rate law for B is approximately 0. Here the individual concentrations of A and C are essentially independent of B.\r\n\r\nAlternatively, if k1 >>> k2, then the rate determining step of the overall process would be k2. Here A would quickly react to form B, but the second reaction would occur over a longer period of time allowing [B] to build up temporarily. In this situation, we can see that B will be an intermediate that can complicate the ability to calculate the differential rate law for C.\r\n\r\nRadioactivity and Half-Lives\r\n\r\nWe can also evaluate the kinetics of a reaction during a half-life. A half-life is the amount of time needed for a reaction to proceed to when there is 1\u20442 the amount of reactant left. For a first order reaction, the half-life can be calculated as\r\n\r\n\ud835\udc611\/2 = ln(2) = 0.693 \ud835\udc58\ud835\udc58\r\n\r\nNote that there is not a variable with the concentration of reactant present. This indicates that for a first order reaction, the half-life is not dependent on the amount of reactant present in the beginning. If we know the rate constant for a reaction, we can determine the half-life.\r\n\r\nFor this experiment, you will be using first order reactions to evaluate the kinetics of a reaction.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n\r\nYou will use your data to calculate the rate constant and half-life of a reaction.\r\n\r\n \r\n\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    <\/div>","rendered":"
    \n
    \n
    \n

    Chesapeake Campus \u2013 Chemistry 112 Laboratory<\/p>\n

    Lab #6- Kinetics: Reaction Rates and Steady State Approximation<\/p>\n

    Objectives<\/p>\n

      \n
    • \uf0b7 \u00a0Determine the rate law for a chemical equation.<\/li>\n
    • \uf0b7 \u00a0Evaluate a two-step reaction with different rate constants using dice.<\/li>\n
    • \uf0b7 \u00a0Determine the rate-determining step of a reaction.<\/li>\n
    • \uf0b7 \u00a0Calculate the rate constant for a reaction.IntroductionKineticsKinetics is the study of how fast a chemical reaction occurs. Some reactions occur very quickly. Consider mixing an aqueous solution of silver nitrate and sodium chloride. As soon as these chemicals touch there is an immediate appearance of the precipitated silver chloride. On the other hand some reactions occur very slowly (even if they are thermodynamically preferred). Technically the decay of diamond to graphite is thermodynamically favored but the reaction has a large activation energy. Due to the activation energy, the reaction takes place extremely slowly leading to the belief that a diamond lasts forever.Rate Law\n

      A rate law is a mathematical description of how fast a chemical reaction takes place. For a given chemical equation<\/li>\n<\/ul>\n<\/div>\n<\/div>\n

      \n
      \n

      the given rate law is defined as<\/p>\n<\/div>\n

      \n

      aA + bB \u2192 cC<\/p>\n

      \ud835\udc5f\ud835\udc4e\ud835\udc61\ud835\udc52 = \ud835\udc58[\ud835\udc34]\ud835\udc65[\ud835\udc35]\ud835\udc66<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      where the rate is defined as being equal to the rate constant (k) multiplied by the concentrations of all reactants raised to the exponents that are different for each chemical reaction. The larger the value of k, generally the faster the reaction occurs. The order of the reaction is equal to the sum of all the exponents. For this experiment you will be working with first order reactions. You will experimentally determine the rate constant for a two distinct situations.<\/p>\n

      There are often reactions that occur in several steps. Consider the situation where a reactant decays into a series of products. We can illustrate this situation with an equation:<\/p>\n

      \ud835\udc581 \ud835\udc582 \ud835\udc4e\ud835\udc34\u2192 \ud835\udc4f\ud835\udc35\u2192 \ud835\udc50\ud835\udc36<\/p>\n

      Here we have a reactant A which decays or reacts to form B. This reaction has the rate constant k1 and can be expressed in terms of the change in concentration of A over time:<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n
      \n

      \ud835\udc51[\ud835\udc34] = \u2212\ud835\udc581[\ud835\udc34] \ud835\udc51\ud835\udc61<\/p>\n

      Here the change in the concentration of A over time is equal to the negative rate constant multiplied by the concentration of A at the given point in time, t. The negative indicates the value of A is being lost as time progresses. The integrated rate law could be written as<\/p>\n

      [\ud835\udc34] = [\ud835\udc34]0 \ud835\udc52\u2212\ud835\udc58\ud835\udc61 \ud835\udc5c\ud835\udc5f ln[\ud835\udc34] = ln[\ud835\udc34]0 \u2212 \ud835\udc58\ud835\udc61<\/p>\n

      In a similar way we can evaluate the second step of the reaction indicated by k2. The best way to do this would be to evaluate the amount of C produced as a function of the rate constant multiplied by the amount of reactant (B) available to react in a given time (t).<\/p>\n

      \ud835\udc51[\ud835\udc36] = \ud835\udc582[\ud835\udc35] \ud835\udc51\ud835\udc61<\/p>\n

      Note that there is no negative which will allow the value of C to go up. To determine the amount of B produced in this chemical process we would have to evaluate BOTH how much B is produced from the first step as well as the amount of B lost in the second step.<\/p>\n

      Rate-Determining Steps<\/p>\n

      Consider the process above. If k2 >>> k1, then as soon as B was created, it would instantly react to make C. The rate determining step of the entire process would be k1 because it would be the slower of the two processes. In addition, because B is used up as quickly as it is made, we can assume the differential-rate law for B is approximately 0. Here the individual concentrations of A and C are essentially independent of B.<\/p>\n

      Alternatively, if k1 >>> k2, then the rate determining step of the overall process would be k2. Here A would quickly react to form B, but the second reaction would occur over a longer period of time allowing [B] to build up temporarily. In this situation, we can see that B will be an intermediate that can complicate the ability to calculate the differential rate law for C.<\/p>\n

      Radioactivity and Half-Lives<\/p>\n

      We can also evaluate the kinetics of a reaction during a half-life. A half-life is the amount of time needed for a reaction to proceed to when there is 1\u20442 the amount of reactant left. For a first order reaction, the half-life can be calculated as<\/p>\n

      \ud835\udc611\/2 = ln(2) = 0.693 \ud835\udc58\ud835\udc58<\/p>\n

      Note that there is not a variable with the concentration of reactant present. This indicates that for a first order reaction, the half-life is not dependent on the amount of reactant present in the beginning. If we know the rate constant for a reaction, we can determine the half-life.<\/p>\n

      For this experiment, you will be using first order reactions to evaluate the kinetics of a reaction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n
      \n

      You will use your data to calculate the rate constant and half-life of a reaction.<\/p>\n

       <\/p>\n

       <\/p>\n<\/div>\n<\/div>\n<\/div>\n

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