For the reaction of H2 with I2, the rate constant is 2.5 × 10-4 dm3 mol-1 s-1 at 327°C and 1.0 dm3 mol-1 s-1 at 527° C. The activation energy for the reaction, in kJ mol-1 is: (R=8.314 J K-1 mol-1)

Question

For the reaction of H2 with I2, the rate constant is 2.5 × 10-4 dm3 mol-1 s-1 at 327°C and 1.0 dm3 mol-1 s-1 at 527° C. The activation energy for the reaction, in kJ mol-1 is: (R=8.314 J K-1 mol-1) (A) 72 (B) 166 (C) 150 (D) 59

Tardigrade Admin · Accepted Answer

H2(g) + I2(g) → 2HI(g) Apply Arrhenius equation log (K2/K1) (Ea/2.303 R) ( (1/600) - (1/800)) log (1/2.5 ×10-4) = (Ea/2.303 ×8.31) ( (200/600 ×800)) ∴ Ea ≈166 kJ/mol

Q. For the reaction of $H_{2}$ with $I_{2}$ , the rate constant is $2.5 \times 1 0^{- 4} d m^{3} m o l^{- 1} s^{- 1}$ at $32 7^{\circ} C$ and $1.0 d m^{3} m o l^{- 1} s^{- 1}$ at $52 7^{\circ} C$ . The activation energy for the reaction, in $k J m o l^{- 1}$ is:
$(R = 8.314 J K^{- 1} m o l^{- 1})$

72

166

150

59

$H_{2} (g) + I_{2} (g) \to 2 H I (g)$
Apply Arrhenius equation
$lo g \frac{K _{2}}{K _{1}} \frac{E _{a}}{2.303 R} (\frac{1}{600} - \frac{1}{800})$
$lo g \frac{1}{2.5 \times 1 0 ^{- 4}} = \frac{E _{a}}{2.303 \times 8.31} (\frac{200}{600 \times 800})$
$∴ E_{a} \approx 166$ kJ/mol

Q. For the reaction of H2​ with I2​, the rate constant is 2.5×10−4dm3mol−1s−1 at 327∘C and 1.0dm3mol−1s−1 at 527∘C. The activation energy for the reaction, in kJmol−1 is: (R=8.314JK−1mol−1)

72

166

150

59

H2​(g)+I2​(g)→2HI(g) Apply Arrhenius equation logK1​K2​​2.303REa​​(6001​−8001​) log2.5×10−41​=2.303×8.31Ea​​(600×800200​) ∴Ea​≈166 kJ/mol

Q. For the reaction of $H_{2}$ with $I_{2}$ , the rate constant is $2.5 \times 1 0^{- 4} d m^{3} m o l^{- 1} s^{- 1}$ at $32 7^{\circ} C$ and $1.0 d m^{3} m o l^{- 1} s^{- 1}$ at $52 7^{\circ} C$ . The activation energy for the reaction, in $k J m o l^{- 1}$ is:
$(R = 8.314 J K^{- 1} m o l^{- 1})$

$H_{2} (g) + I_{2} (g) \to 2 H I (g)$
Apply Arrhenius equation
$lo g \frac{K _{2}}{K _{1}} \frac{E _{a}}{2.303 R} (\frac{1}{600} - \frac{1}{800})$
$lo g \frac{1}{2.5 \times 1 0 ^{- 4}} = \frac{E _{a}}{2.303 \times 8.31} (\frac{200}{600 \times 800})$
$∴ E_{a} \approx 166$ kJ/mol