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kkurt [141]
3 years ago
9

Write a balanced half-reaction for the reduction of gaseous nitrogen (N2) to aqueous hydrazine (N2H4) in basic aqueous solution.

Chemistry
1 answer:
weqwewe [10]3 years ago
5 0

Answer:

N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)

Explanation:

The half reaction for the reaction for the reduction of gaseous nitrogen to aqueous hydrazine is;

N₂(g) → N₂H₄(aq)

The balancing the atoms in the half reaction. Hydrogen atom is balanced by adding hydrogen ions (H⁺)

We have;

N₂(g) + 4H⁺(aq) → N₂H₄(aq)

Then we balance the charge on both sides by adding electrons where the positive charge is greater;

we have;

N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)

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Answer:

By increasing the pressure, the molar concentration of  N2O4 will increase

Explanation:

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This equation is reversible and exotherm. By <u>decreasing the temperature</u>, the reaction will produce more energy, so the reaction will move to the right.  But a lower temperature also lowers the rate of the process, so, the temperature is set at a compromise value that allows N2O4 to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable

So <u>increasing the temperature</u> will shift the equilibrium to the left. The equilibrium shifts in the direction that consumes energy.

If we d<u>ecrease the concentration of NO2</u>, the equilibrium will shift to the left, resulting in forming more reactants.

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By increasing the pressure, the molar concentration of  N2O4 will increase

7 0
3 years ago
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Which substance produces hydroxide ions in solution?
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Answer:

An Arrhenius Base

Explanation:

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8 0
2 years ago
Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
3 years ago
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Explanation:

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The balanced equation is attached in the image below. The coefficients are 2, 2, blank.

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