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Mumz [18]
3 years ago
12

Write and Balance: Exothermic reaction between aqueous sodium hydroxide and sulphuric acid.

Chemistry
1 answer:
kolezko [41]3 years ago
7 0
Exothermic reaction discharges energy in forms of light, heat or sound, thus it is represented by negative flow of heat. In this equation:
<span><span>
NaOH (aq) + H2SO4 (aq) </span>à<span> H2O (l) + Na2SO4 (aq)</span></span>
<span>
When the solution of Sodium hydroxide is added to the Sulphuric acid, water and Sodium sulfate is produced. To balance this, we put (2) two on Sodium hydroxide and (2) two on water.</span>
<span><u>
2</u><span>NaOH (aq) + H2SO4 (aq) </span>à<u> 2</u><span>H2O (l) + Na2SO4 (aq)</span></span> <span> </span>
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The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

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What is the reaction order in H_2?

What is the reaction order in I_2?

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At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

So when

      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

3 0
2 years ago
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