For the following reaction at $0^oC$: $H_2SeO_3$ $+$ $6I^-$ $+$ $4H^+$ $\longrightarrow$ $Se$ $+$ $2I_3^-$ $+$ $3H_2O$ - the rate constant is $k$ $=$ $5.0\cdot 10^5\frac{L^5}{mol^5\cdot s}$ - the order . with respect to $H_2SeO_3$ is $1$ . with respect to $I^-$ is $3$ . with respect to $H^+$ is $2$ Calculate the initial rate, knowing at the beginning: $[H_2SeO_3]$ $=$ $2.0\cdot 10^{-2}\cdot M$ $[I^-]$ $=$ $2.0\cdot 10^{-3}\cdot M$ $[H^+]$ $=$ $1.0\cdot 10^{-3}\cdot M$
