Question
In a reaction, $$2HI \to {H_2} + {I_2},$$ the concentration of $$HI$$ decreases from $$0.5\,mol\,{L^{ - 1}}$$ to $$0.4\,mol\,{L^{ - 1}}$$ in 10 minutes. What is the rate of reaction during this interval?
A.
\[5\times {{10}^{-3}}\,M\,{{\min }^{-1}}\]
B.
\[2.5\times {{10}^{-3}}\,M\,{{\min }^{-1}}\]
C.
\[5\times {{10}^{-2}}\,M\,{{\min }^{-1}}\]
D.
\[2.5\times {{10}^{-2}}M\,{{\min }^{-1}}\]
Answer :
\[5\times {{10}^{-3}}\,M\,{{\min }^{-1}}\]
Solution :
\[\begin{align}
& \text{Average rate} \\
& =-\frac{1}{2}\frac{\Delta \left[ HI \right]}{\Delta t} \\
& =-\frac{1}{2}\times \frac{0.4-0.5}{10} \\
& =\frac{1}{2}\times \frac{0.1}{10} \\
& =5\times {{10}^{-3}}M\,{{\min }^{-1}} \\
\end{align}\]