The Arrhenius equation is
so if we plot a graph between lnK (natural log of rate constant) and 1/T (inverse of temperature in kelvin)
The straight line equation will be
lnK = lnA - Ea/RT
Where
A = constant for a particular reaction at any temperature
R = gas constant = 8.314 J /molK
So as per this equation the slope of line will be = -Ea/R
Or
Ea = -slope X R
Ea = -(-9.20X10³X 8.314) =76.49kJ /mol
Answer: activation energy = 76.49 kJ /mol
Answer:
pH of the benzoic acid solution = 3.35
Explanation:
Benzoic acid is a weak monoprotic acid
pKa = 4.2
⇒ Ka = 6.3 x 10⁻⁵
C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺
Initial concentration C molar 0 0
At equilibrium (c - cα) cα cα
Acid dissociation constant Ka =
⇒ 6.3 x 10⁻⁵ = α²c ................(1)
(for weak electrolyte 1 -α = 1)
Concentration of benzoic acid C = 25 x 0.126 x 10⁻³ = 0.00315 molar
From eqn (1)
α² =
⇒ α = √( )
⇒ α = 0.14
So concentration of H⁺ ion = 0.14 x 0.00315 = 0.000441
pH = - log 0.000441 = 3.35
We not considered H⁺ concentration of water because pH < 7