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
Answer:
Explanation:
Answer:
Its a liquid cause it flows
Explanation:
think about it none of the others make sense ice is a solid and its clear
something like a wall can be thick but it doesn't flow
Answer:
Changing the surface area from minimum to maximum increases the number of reactants in a chemical reaction.
An increase in surface area of a solid reactant means more of its particles are exposed to attack by the other Particle. This results in an increased chance of collisions between reactant particles, so there are more collisions in any given time and the rate of reaction increases.
Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
Answer:
107.8
Explanation:
64 gram of N2H4 produce 72 gram of H20
then by crossmultiplication
64*121.3/72=107.82
