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Standard XII

Chemistry

Stoichiometric Calculations

When 50 ml of...

Question

When 50 ml of 2 molar so has of $[N_{2} O_{5}]$ was heated 0.28 L of $O_{2}$ at NTP was formed after 30 min. Calculate the concentration of unreacted $N_{2} O_{4}$ at that time and also the average rate.
$N_{2} O_{5} \to N_{2} O_{4} + \frac{1}{2} O_{2}$

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Solution

$N_{2} O_{5} \to N_{2} O_{4} + 1 / 2 O_{2}$
$50 m^{2} 2 M m = \frac{n}{v} \Rightarrow n = m \times v$
$= 2 (\frac{50}{1000}) = 0.1$
1 moles $N_{2} O_{5}$ reacts to give 1 mole $O_{2}$
i.e : $n o_{2} = \frac{0.28}{22.4} = 0.0125 m o l e O_{2}$
$0.0125 m o l e N_{2} O_{5}$ use
Remaining = $(0.1 - 0.0125)$ mol
$= 0.0875$ mol

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When 50 ml of 2M solution of N2O5 was heated, 0.28L of O2 at NTP was fgormed after 30 min. Cal the conc of unreacted N2O5 at that time and find rate of reaction.

Q. The composition of

N_{2} O_{5}

is first order reaction represented by:

N_{2} O_{5} \to N_{2} O_{4} + 1 / 2 O_{2} .

After 20 min the volume of

O_{2}

poduced is 10 mL and at the end of the reaction 40 mL. The rate constant is equal to:

Q. The half life of decomposition of

N_{2} O_{5}

is a first order reaction represented by

N_{2} O_{5} \to N_{2} O_{4} + 1 / 2 O_{2}

After 15 min the volume of

O_{2}

produced is 9mL and at the end of the reaction 35 mL. The rate constant is equal to:

N_{2} O_{5}

decomposes to

N_{2} O_{4}

and

O_{2}

as :

N_{2} O_{5} \to N_{2} O_{4} + \frac{1}{2} O_{2}

The pressure

P_{1}

at any stage is related to

p_{0}

and

x

, the fraction of dissociation, as:

Q. When

N_{2} O_{5}

is heated at temperature T, it dissociates as

N_{2} O_{5} ⇌ N_{2} O_{3} + O_{2}, K_{e} = 2.5

. At the time

N_{2} O_{3}

also decomposes as:

N_{2} O_{3} ⇌ N_{2} O + O^{2}

. If initially

4

moles of

N_{2} O_{5}

are taken in

1

litre flask and allowed to attain equilibrium, concentration of

O_{2}

was formed to be

2.5 M

. Equilibrium concentration of

N_{2} O

is:

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When 50 ml of 2 molar so has of [N2O5] was heated 0.28 L of O2 at NTP was formed after 30 min. Calculate the concentration of unreacted N2O4 at that time and also the average rate. N2O5→N2O4+12O2

N2O5→N2O4+1/2O250m22Mm=nv⇒n=m×v=2(501000)=0.11 moles N2O5 reacts to give 1 mole O2i.e : no2=0.2822.4=0.0125moleO20.0125moleN2O5 useRemaining = (0.1−0.0125) mol=0.0875 mol

When 50 ml of 2 molar so has of $[N_{2} O_{5}]$ was heated 0.28 L of $O_{2}$ at NTP was formed after 30 min. Calculate the concentration of unreacted $N_{2} O_{4}$ at that time and also the average rate.
$N_{2} O_{5} \to N_{2} O_{4} + \frac{1}{2} O_{2}$