For the reaction HO(g)+ H₂(g)$\rightarrow$H₂O(g)+ H(g) A plot of $\ln k$ Versus 1/T gives a straight line with a slope equal to -5.1 * 10³ K.What is the activation energy for the reaction? A) 42 kJ.mol^-¹ B) 98 kJ.mol^-¹ C) 0.61 kJ.mol^-¹ D) 5.1 kJ.mol^-¹ E) 12 kJ.mol^-¹

Question

Quizplus · Accepted Answer

The slope of the plot of $\ln k$ versus 1/T is equal to $-E_a/R$, where $E_a$ is the activation energy and $R$ is the gas constant (8.314 J/mol·K). Solving for $E_a$ gives $E_a = -\text{slope} \times R = 5.1 \times 10^3 \times 8.314 = 42,416.4$ J/mol, which is approximately 42 kJ/mol.

For the Reaction HO(g)+ H2(g) →\rightarrow→ H2O(g)+ H(g) a Plot Of ln⁡k\ln klnk

For the Reaction
HO(g)+ H₂(g) $\rightarrow$ H₂O(g)+ H(g)
a Plot Of $\ln k$