For the reaction of H2 with I2 , the rate constant is 2.5×10−4dm3mol−1s−1 at 327°C and 1.0dm3mol−1s−1527∘C. The activation energy for the reaction, in kJ mol-1 is R=8.314Jk−1mol−1

H2(g)+I2(g)⟶2HI(g)log⁡K2K1=Ea2.303R1600−1800log⁡12.5×10−4=Ea2.303×8.314200600×800Ea=166kJ/mol

For the reaction of H2 with I2 , the rate constant is 2.5×10−4dm3mol−1s−1 at 327°C and 1.0dm3mol−1s−1527∘C. The activation energy for the reaction, in kJ mol-1 is R=8.314Jk−1mol−1

H2(g)+I2(g)⟶2HI(g)log⁡K2K1=Ea2.303R1600−1800log⁡12.5×10−4=Ea2.303×8.314200600×800Ea=166kJ/mol

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For the reaction of H₂ with I₂ , the rate constant is $2 .5 \times 10^{- 4} {dm}^{3} {mol}^{- 1} s^{- 1}$ at 327°C and $1 .0 {dm}^{3} {mol}^{- 1} s^{- 1} 527^{\circ} C$ . The activation energy for the reaction, in kJ mol^-1 is $(R = 8 .314 {Jk}^{- 1} {mol}^{- 1})$

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Detailed Solution

$\begin{array}{l} H_{2 (g)} + I_{2 (g)} ⟶ 2 {HI}_{(g)} \log (\frac{K_{2}}{K_{1}}) = \frac{Ea}{2 .303 R} [\frac{1}{600} - \frac{1}{800}] \\ \log (\frac{1}{2 .5 \times 10^{- 4}}) = \frac{Ea}{2 .303 \times 8 .314} [\frac{200}{600 \times 800}] E_{a} = 166 kJ / mol \end{array}$