Chemistry, 27.05.2020 19:06 FriendlyDude640

Rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is the reaction order in H2? What is the reaction order in I2? 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.

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Answer from: ale1910

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.

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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Answer from: Quest

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o

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