Question

Thermal decomposition for n2o5(g)n2o5(g) follows 1st order kinetics as per n2o5(g)→2no(g)+12o2(g)n2o5(g)→2no(g)+12o2(g). Initial pressure of n2o5n2o5 is 100 mm. After 10 minutes, the total pressure developed is 130 mm. What is rate constant ? (take log2 = 0.3)

Answer 1

Explanation:

brainlest please.......

Answer 2

Answer:

Value of rate constant will be$9.212 \times 10^{-3} \mathrm{~min}^{-1}$

Explanation:

Hint: In first-order kinetics at any point rate of reaction is directly proportional to the concentration of reactants and the concentration of products formed totally depends on how many reactants have reacted and what is the stoichiometry of the reaction. Now as we know the pressure of a gaseous substance is directly proportional to its pressure at a constant temperature, therefore the ratio of change in concentrations will always equal the ratio of change in their partial pressures.

Complete step by step answer:

$N_{2}O_{5} === > NO_{2}+12O_{2}$

According to reaction stoichiometry, when 1 mole of N2O5 decomposes, it forms 1 mole NO2 and half moleO2.

We also know that at a given temperature for constant volume, pressure will always be proportional to the number of moles of the gas; therefore change in pressure will also be proportional to the change in the number of moles.

- So, we can say that according to reaction stoichiometry, if pressure of N2O5 decreases by x unit then pressure of NO2 will be x unit and pressure of oxygen gas will be x2 units.

In the system,

Partial pressure of $N_{2}O_{5} = (100−x)mm$

→ Partial pressure of $NO_{2} = x mm$

→ Partial pressure of O2= x2 mm

- According to dalton’s partial pressure law,

Total pressure of the system will be equal to the sum of the partial pressure of all gases.

$P=pN_{2}O_{5}+pNO_{2}+pO_{2}$

$130=100-x+x+\frac{x}{2}$

$== > 130=100+\frac{x}{2}$

$== > 130-100=\frac{x}{2}$

$== > \frac{x}{2} =30$

$x=30*2\\x=60mm$

- After 10 minutes

→ Pressure of $N_{2}O_{5}$ $= 100 - 60$

→ Pressure of $N_{2}O_{5}$ $= 40 mm$

- Now let’s use first order integrated rate equation to find out the value of rate constant

$k=\frac{2.303}{t} log\frac{p}{p_{0}}$

Here, p is the pressure of reactant at time t and p0 is the pressure of the reactant at the start of the reaction.

$p_{0}=100mm\\p=40mm\\t=10min$

- Now put all these values into the integrated rate equation

$\rightarrow k=\frac{2.303}{100} \log \frac{100}{40} \\ \rightarrow k=\frac{2.303}{100} \log 2.5 \\ \rightarrow k=\frac{2.303}{100} \times 0.4 \\ \rightarrow k=\frac{0.9212}{100} \\ \rightarrow k=9.212 \times 10^{-3} \mathrm{~min}^{-1}$

Therefore value of rate constant will be$9.212 \times 10^{-3} \mathrm{~min}^{-1}$