Thermal decomposition for N2O5(g) 1st order kinetics as per N2O5(g)→2NO2(g)+12O2(g) Initial presure of N2O5 is 100mm. After 10 minutes, the pressure developed is 130mm, what is rate constant?

N2O5t =100 mm→2NO2+12O2 t=10 min 100-x 2x x2at t= 10 mins 100-x+2x+x2=130x = 20K1=2.30310log100100-20=0.2303(1-0.9010)=0.02303 min-1

Thermal decomposition for N2O5(g) 1st order kinetics as per N2O5(g)→2NO2(g)+12O2(g) Initial presure of N2O5 is 100mm. After 10 minutes, the pressure developed is 130mm, what is rate constant?

N2O5t =100 mm→2NO2+12O2 t=10 min 100-x 2x x2at t= 10 mins 100-x+2x+x2=130x = 20K1=2.30310log100100-20=0.2303(1-0.9010)=0.02303 min-1

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Thermal decomposition for $N_{2} O_{5 (g)}$ 1^storder kinetics as per $N_{2} O_{5 (g)} \to 2 {NO}_{2 (g)} +$ $\frac{1}{2} O_{2 (g)}$ Initial presure of $N_{2} O_{5}$ is 100mm. After 10 minutes, the pressure developed is 130mm, what is rate constant?

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0.693 min^–1

$2.303 \times 10^{- 2}$ min^-1

0.03 min^–1

0.025 min^–1

answer is C.

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Detailed Solution

$\overset{t = 100 mm}{N_{2} O_{5}} \to 2 N O_{2} + \frac{1}{2} O_{2} t = 10 m i n 100 - x 2 x \frac{x}{2}$

at t= 10 mins

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

x = 20

$K_{1} = \frac{2.303}{10} \log \frac{100}{100 - 20} = 0.2303 (1 - 0.9010) = 0.02303 \min^{- 1}$

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