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Fudgin [204]
2 years ago
11

Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise

to 0.150 atm .
Chemistry
1 answer:
USPshnik [31]2 years ago
3 0

Answer:

t = 37.1 s

Explanation:

The equation for the reaction is given as;

                  2 N2O5(g)  --> 4 NO2 + O2

Initial:          0.110                  -             -

change:        -2x                  +4x        +x

Final:          0.110 - 2x           +4x        +x

But final = 0.150atm;

0.110 - 2x    +  4x   +  x = 0.150 atm

3x = 0.150 - 0.110

x = 0.0133 atm

Pressure in reactant side;

0.110 - 2x

0.110 - 2 (0.0133) = 0.0834 atm

The integral rate law expression is given as;

ln ( [A] / [Ao] ) = -kt

k =  rate constant = 7.48*10^-3*s-1

ln (0.0834/0.11) = (7.48*10^-3)  t

upon solving, t = 37.1 s

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1 year ago
Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution
Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

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3 years ago
The standard free energy of activation of one reaction A is 95.00 kJ mol–1 (22.71 kcal mol–1). The standard free energy of activ
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Answer:

The answer to the questions are as follows

Reaction B is 4426.28 times faster than reaction A

(b) Reaction B is faster.

Explanation:

To solve the question we are meant to compare both reactions to see which one is faster

The values of the given activation energies are as follows

For A

Ea = 95.00 kJ mol–1 (22.71 kcal mol–1) and

for  B

Ea = 74.20 kJ mol–1 (17.73 kcal mol–1)

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The Arrhenius Law is given by

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Where

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Therefore

For reaction A, the rate constant k₁ is given by k₁ = Ae^{\frac{-95000}{(8.314)(298)} }

And for B the rate constant k₂ is given by k₂ = Ae^{\frac{-74200 }{(8.314)(298)} }

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