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CaHeK987 [17]
3 years ago
7

need help plz !The reaction between hydrogen and nitrogen to give ammonium is a reversible reaction. if why or why not.Answer it

plz ​don't spam
Chemistry
2 answers:
Orlov [11]3 years ago
7 0

Answer:

<h3 /><h3>We get ammonia because the </h3><h3>forward and reverse reactions are happening at the same rates.</h3>

<h3>If 3 mol of H2 is mixed in a sealed vessel with 1 mol N2 under suitable conditions thenthey will react to form ammonia NH3 :</h3>

<h3>N2 + 3H2→ 2NH3</h3>

<h3>At the start of the reaction the concentration</h3><h3> of the </h3><h3>N2 and H2 are high. As soon as </h3><h3>some NH3 is formed the reverse reaction will start to occur:</h3><h3 /><h3>2NH3→N2+3H2</h3>

  • The rate of the reaction depends on concentration so the forward reaction will be fast at first when the concentration of the reactants is high. It will slow down as their concentration decreases.

  • By the same reasoning the reverse reaction will be slow at first then increase. These two processes continue until a point is reached when the rates of the forward and reverse reactions are equal.

  • We now state that the reaction has reached equilibrium which we show by:

  • N2+3H2⇌2NH3

It is described as "dynamic" because the forward and reverse reactions are happening at the same time although the concentrations of all the species remain constant.

So although NH3 is constantly breaking down, more is being formed at a constant rate.

In the Haber Process the system is actually not allowed to completely reach equilibrium as the process is continuous, as described in Mukhtar's answer.

MaRussiya [10]3 years ago
5 0

Answer:

We get ammonia because the forward and reverse reactions are happening at the same rates.

If 3 mol of  

H

2

is mixed in a sealed vessel with 1 mol  

N

2

under suitable conditions then they will react to form ammonia  

N

H

3

:

N

2

+

3

H

2

→

2

N

H

3

At the start of the reaction the concentration of the  

N

2

and  

H

2

are high. As soon as some  

N

H

3

is formed the reverse reaction will start to occur:

2

N

H

3

→

N

2

+

3

H

2

The rate of the reaction depends on concentration so the forward reaction will be fast at first when the concentration of the reactants is high. It will slow down as their concentration decreases.

By the same reasoning the reverse reaction will be slow at first then increase. These two processes continue until a point is reached when the rates of the forward and reverse reactions are equal.

We now state that the reaction has reached equilibrium which we show by:

N

2

+

3

H

2

⇌

2

N

H

3

It is described as "dynamic" because the forward and reverse reactions are happening at the same time although the concentrations of all the species remain constant.

So although  

N

H

3

is constantly breaking down, more is being formed at a constant rate.

In the Haber Process the system is actually not allowed to completely reach equilibrium as the process is continuous, as described in Mukhtar's answer.

Explanation:

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Answer:

Value of Q_{p} for the given redox reaction is 1.0\times 10^{-8}

Explanation:

Redox reaction with states of species:

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Reaction quotient for this redox reaction:

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Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of H_{2}O is taken as 1 due to the fact that H_{2}O is a pure liquid.

pH=-log[H^{+}]

So, [H^{+}]=10^{-pH}

Plug in all the given values in the equation of Q_{p}:

Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}

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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

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