Experiments show that if the chemical reaction N 2 O 5 → 2 N O 2 + 1 2 O 2 takes place at 45 ∘ C, the rate of the reaction of dinitrogen pentoxide is proportional to its concentration as follows: − d [ N 2 O 5 ] d t = 0.0005 [ N 2 O 5 ]a) Find an expression for the concentration [ N 2 O 5 ] after t seconds if the initial concentration is C.b) How long will the reaction take to reduce the concentration of N 2 O 5 to 90% of its original value?Part aI simply used the instantaneous rate of 0.0005 as the constant k (referring to the rate of reaction) and, since C is a constant, I utilized it as the initial value in the formula: y ( t ) = y ( o ) e k t = C e k t I replaced k for 0.0005: the book has a negative sign in front of this value, I am curious as to why since the equal sign would indicate otherwise, no? Nonetheless, the equation: C e − 0.0005 ( t ) Please comment on my reasoning.Part bI simply utilized the above equation and set it equal to the decimal form of 90% with a negative sign since the question said reduced C e − 0.0005 ( t ) = − 0.9However, I really don't understand conceptually why 90% can be considered 0.9 of the concentration. I just need to clear this up in my mind.From that point I am pretty lost why the C appears on both sides of the equation and why it disappears.Answer from the book: y ( t ) = C e − 0.0005 = 0.9 C → C e − 0.0005 = 0.9 C → e − 0.0005 = 0.9What happened to C?

Question

Experiments show that if the chemical reaction            N      2              O      5        →    2        N          O      2        +          1      2                  O      2      takes place at   45          ∘    C, the rate of the reaction of dinitrogen pentoxide is proportional to its concentration as follows:  −            d      [                        N          2                          O          5                    ]              d      t        =  0.0005  [            N      2              O      5        ]a) Find an expression for the concentration [            N      2              O      5      ] after t seconds if the initial concentration is C.b) How long will the reaction take to reduce the concentration of             N      2              O      5       to 90% of its original value?Part aI simply used the instantaneous rate of 0.0005 as the constant k (referring to the rate of reaction) and, since C is a constant, I utilized it as the initial value in the formula:  y  (  t  )  =  y  (  o  )      e          k      t        =  C      e          k      t      I replaced k for 0.0005: the book has a negative sign in front of this value, I am curious as to why since the equal sign would indicate otherwise, no? Nonetheless, the equation:  C      e          −      0.0005      (      t      )      Please comment on my reasoning.Part bI simply utilized the above equation and set it equal to the decimal form of 90% with a negative sign since the question said reduced  C      e          −      0.0005      (      t      )        =  −  0.9However, I really don&#039;t understand conceptually why 90% can be considered 0.9 of the concentration. I just need to clear this up in my mind.From that point I am pretty lost why the C appears on both sides of the equation and why it disappears.Answer from the book:  y  (  t  )  =  C      e          −      0.0005        =  0.9  C  →  C      e          −      0.0005        =  0.9  C  →      e          −      0.0005        =  0.9What happened to C?

bewagox7 · Accepted Answer

First of all, we denote the concentration of             N      2              O      5       at time t with y(t). Then your governing equation is  −                    d        y        (        t        )                    d        t              =  k  y  (  t  )The solution to this simple ordinary differential equation (ODE) is  y  (  t  )  =  A            e              −        k        t            where A is an arbitrary constant. You can simply check this by substituting into the ODE. Your first mistake was to take   y  (  t  )  =  A            e              k        t             as the solution of the ODE, neglecting the minus sign. Now, we have the initial condition  y  (  0  )  =  Cwhere C is a known constant which is the initial concentration of             N      2              O      5      . Then our solution becomes  y  (  t  )  =  C            e              −        k        t            So the solution of Part a is finished since we have found the concentration of             N      2              O      5       at time t in terms of its initial concentration. Part b wants to know at what time the concentration is 90% of the initial concentration, i.e.  y  (            t      ∗        )  =                    90                    100              y  (  0  )  =                    90                    100              C  =  0.9  Cwhere t∗ is the time we want to find. So we may write  C            e              −        k                              t            ∗                                =  0.9  CDivide the above equation by C            e              −        k                              t            ∗                                =  0.9taking natural logarithm from both sides we have  −  k            t      ∗        =  ln  ⁡  (  0.9  )and finally, solving for t∗ leads to            t      ∗        =  −                    ln        ⁡        (        0.9        )            k

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