The correct answer is (B) : The velocity of atomic oxygen doubles
As
\(v_{rms}=\sqrt{\frac{3RT}{M_0}}\)
T is doubled and oxygen molecule is dissociated into atomic oxygen molar mass is halved.
So,
\(v'_{rms}=\sqrt{\frac{3R×2T_0}{M_0/2}}=2v_{rms}\)
So velocity of atomic oxygen is doubled.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).