In the reaction $ 2N_2O_5 \rightarrow 4NO_2 + O_2 , initial pressure is 500 atm and rate constant K is 3.38 10^{-5} sec^{-1} . After 10 minutes the final pressure of N_2O_5 $ is
490 atm , $ K = { 2.303 \over t} log { Po \over Pt } \therefore 3.38 \times 10 ^ {-5} = { 2.303 \over 600 } log { 500 \over Pt } $ $ log { 500 \over Pt } = 0.0088 OR { 500 \over pt } = 1.021 OR pt = 490 atm $