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Calorimetry

In a reaction...

Question

In a reaction; $H_{2} + I_{2} \to 2 H I$ , the rate of dissappearance of $I_{2}$ is found to be $10^{- 6} m o l l i t r e^{- 1} s e c^{- 1}$ . If the corresponding rate of appearance of HI is $x \times 10^{- 6} m o l l i t r e^{- 1} s e c^{- 1}$ then $x$ is:

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Solution

The rate of appearance of HI is twice the rate of disappearance of $I_{2}$ .
It is equal to $2 \times 10^{- 6} m o l / L / s e c = x \times 10^{- 6} m o l / L / s e c$
Hence, $x = 2$ .

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