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3 years ago

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Which of the molecules can be involved in hydrogen bonding? HF CH3OH CH4 H2

1 answer:

Svetllana [295]3 years ago

4 0

Answer:

HF & CH3OH

Explanation:

Hydrogen bonds are formed as a consequence of the attractions between a partially positive hydrogen (δ+ hydrogen) atom of a molecule and a lone pair of another electronegative element, such as a nitrogen, oxygen or fluorine molecule of another molecule.

In the case of HF, the partially positive hydrogen atom makes a hydrogen bond with the lone pair of a fluorine atom of another HF molecule in its proximity.

In the case of CH3OH, the partially positive hydrogen atom (bound to Oxygen) makes a hydrogen bond with the lone pair of an oxygen atom of another CH3OH molecule in its proximity.

<h3>Hope this help!</h3>

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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 $H_2$ ?

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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

$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

$k = 35.23 M^{-2} s^{-1}$

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create dates 2004-09-16

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