Answer:
k₂ = 4.06 x 10⁻² s⁻¹.
Explanation:
- From Arrhenius law: <em>K = Ae(-Ea/RT)</em>
where, K is the rate constant of the reaction.
A is the Arrhenius factor.
Ea is the activation energy.
R is the general gas constant.
T is the temperature.
- At different temperatures:
<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>
k₁ = 5.8 × 10⁻³ s⁻¹, k₂ = ??? , Ea = 33600 J/mol, R = 8.314 J/mol.K, T₁ = 298.0 K, T₂ = 348.0 K.
- ln(k₂/5.8 × 10⁻³ s⁻¹) = (33600 J/mol / 8.314 J/mol.K) [(348.0 K - 298.0 K) / (298.0 K x 348.0 K)] = (4041.37) (4.82 x 10⁻⁴) = 1.9479.
- Taking exponential of both sides:
(k₂/5.8 × 10⁻³ s⁻¹) = 7.014.
∴ k₂ = 4.06 x 10⁻² s⁻¹.
Answer:- d. -0.35 degree C is the right answer.
Solution:- depression in freezing point, delta Tf formula is...
delta Tf = i x m x kf
where, i is Van't hoff factor. glucose is covalent compound and does not dissociate to give ions and so the value of i for this is 1.
m is molaity and kf is the molal freezing point constant.
Let's plug in the given values in the formula..
delta Tf = 1 x 0.19 x 1.858
delta Tf = 0.35
Pure water freezes at 0 degree C.
So, freezing point of solution = 0 - 0.35 = -0.35 degree C
There are 2 grams remain
<h3>Further explanation</h3>
Given
t1/2 = 30 years
t = 170 years
No = 100 g
Required
Remaining sample
Solution
General formulas used in decay:
Input the value :
