In the reaction 2N_2O_5 to 4NO_2 + O_2,the in | vedclass

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(A) For a first-order reaction,the rate constant $K$ is given by the formula: $K = \frac{2.303}{t} \log_{10} \frac{p_0}{p_t}$.Given: $p_0 = 500 \ atm$

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$490$

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(A) For a first-order reaction,the rate constant $K$ is given by the formula: $K = \frac{2.303}{t} \log_{10} \frac{p_0}{p_t}$.Given: $p_0 = 500 \ atm$,$K = 3.38 	imes 10^{-5} \ s^{-1}$,and $t = 10 \ minutes = 600 \ s$.Substituting the values: $3.38 	imes 10^{-5} = \frac{2.303}{600} \log_{10} \frac{500}{p_t}$.$0.00880 = \log_{10} \frac{500}{p_t}$.Taking the antilog: $10^{0.00880} = \frac{500}{p_t} \approx 1.0204$.$p_t = \frac{500}{1.0204} \approx 490 \ atm$.

In the reaction $2N_2O_5 \to 4NO_2 + O_2$,the initial pressure is $500 \ atm$ and the rate constant $K$ is $3.38 \times 10^{-5} \ s^{-1}$. After $10 \ minutes$,the final pressure of $N_2O_5$ is ........ $atm$.

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