Answer:
<h2>91.2</h2>
Explanation:
Given that:
Temperature = 60 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (60 + 273.15) K = 333.15 K
n = 5.14 × 10⁻³ mol
V = 1,171 mL = 1.171 L ( 1 mL = 0.001 L)
Using ideal gas equation as:

where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.364 L Torr/ K mol
Applying the equation as:
P × 1.171 L = 5.14 × 10⁻³ mol × 62.364 L Torr/ K mol × 333.15 K
<u>⇒P = 91.2 torr</u>
Answer:
cetacean i think thats the answer
Answer:
The order of reaction is 2.
Rate constant is 0.0328 (M s)⁻¹
Explanation:
The rate of a reaction is inversely proportional to the time taken for the reaction.
As we are decreasing the concentration of the reactant the half life is increasing.
a) For zero order reaction: the half life is directly proportional to initial concentration of reactant
b) for first order reaction: the half life is independent of the initial concentration.
c) higher order reaction: The relation between half life and rate of reaction is:
Rate = ![\frac{1}{k[A_{0}]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%7D)
Half life =![K\frac{1}{[A_{0}]^{(n-1)} }](https://tex.z-dn.net/?f=K%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%20%7D)
![\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }](https://tex.z-dn.net/?f=%5Cfrac%7B%28halflife_%7B1%7D%29%7D%7B%28halflife_%7B2%7D%29%7D%3D%5Cfrac%7B%5BA_%7B2%7D%5D%5E%7B%28n-1%29%7D%7D%7B%5BA_%7B1%7D%5D%5E%7B%28n-1%29%7D%20%7D)
where n = order of reaction
Putting values
![\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B109%7D%7B231%7D%3D%5Cfrac%7B%5B0.132%5D%5E%7B%28n-1%29%7D%7D%7B%5B0.280%5D%5E%7B%28n-1%29%7D%7D)

Hence n = 2
![halflife=\frac{1}{k[A_{0}]}](https://tex.z-dn.net/?f=halflife%3D%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%7D)
Putting values

K = 0.0328