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postnew [5]
3 years ago
10

The equilibrium: 2 NO2(g) \Longleftrightarrow⇔ N2O4(g) has Kc = 4.7 at 100ºC. What is true about the rates of the forwar

d (ratefor) and reverse (raterev) reactions initially and at equilibrium if an empty container is filled with just NO2?
Initial: forward rate < reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rate
Initial: forward rate = reverse rate Equilibrium: forward rate = reverse rate
Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

  • Initial: forward rate > reverse rate
  • Equilibrium: forward rate = reverse rate

Explanation:

2NO₂(g) → N₂O₄(g)   Kc=4.7

The definition of <em>equilibrium</em> is when the forward rate and the reverse rate are <em>equal</em>.

Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.

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S be x.

Now,

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<h3>What is oxidation number?</h3>

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1 year ago
Two plates slide past each other in opposite directions at a ____ boundary.
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When an ionic compound dissolves in water: When an ionic compound dissolves in water: the solvent-solute attractive forces overc
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Answer:

each of the above (A, B, and C) occurs

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--  the water dipoles' positive end attracts the negative ions

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2 years ago
rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
Sunny_sXe [5.5K]

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}

     

     

3 0
2 years ago
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