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icang [17]
2 years ago
5

A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon.

Chemistry
1 answer:
zalisa [80]2 years ago
4 0

Answer:

a) Same

b) Nitrogen

c) Same

d) Nitrogen

Explanation:

a)

The formula for partial pressure of a gas is equal to

p_B = n_B \frac{RT}{V}

Here nB is the number of moles .

The number of moles for both the gases are same and hence the partial pressure for the two gases will also be same.

b) The greater average velocity is calculated by using following formula

v_{RMS} = \sqrt{3RTM}

Here M is the molar mass.

Molar mass of nitrogen is greater than the molar mass of xenon and hence nitrogen will have higher greater average velocity

c) As we know, the average kinetic energy of gas particles is dependent on the absolute temperature of gas and if all the gases are at same temperature, their kinetic energy will also be same. Since nitrogen and xenon are at same temperature, their kinetic energy will be same

d) Effusivity is depended directly on  the thermal conductivity,  density and  and the specific heat capacity.

All these three parameters are higher in case of nitrogen. Thus, it will effuse first

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rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
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Complete Question

The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

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}

     

     

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