## Chemistry: Principles and Practice (3rd Edition)

Rate of disappearance of $NO_{2}$= 0.011 M/s. Rate of the reaction= 0.0055 M/s.
Rate of the reaction= $\frac{\Delta [N_{2}O_{5}]}{\Delta t}=-\frac{1}{2}\times\frac{\Delta [NO_{2}]}{\Delta t}= 0.0055 M/s$ $\implies \frac{\Delta [NO_{2}]}{\Delta t}= -2\times\frac{\Delta [N_{2}O_{5}]}{\Delta t}$ $=-2\times0.0055M/s= -0.011 M/s$. Negative sign indicates the disappearance and the rate of disappearance of $NO_{2}$ is 0.011 M/s.