Answer:
A) t = 22.5 min and B) t = 29.94 min
Explanation:
Initial concentration, [A]₀ = 100
Final concentration = 100 -75 = 25
Time = 45 min
A) First order reaction
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[25] − ln[100] = - 45k
-1.386 = -45k
k = 0.0308 min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[50] − ln[100] = - 0.0308 * t
-0.693 = -0.0308 * t
t = 22.5 min
B) Zero Order
[A] = [A]₀ − kt
Using the values from the initial reaction and solving for k, we have;
25 = 100 - k(45)
-75 = -45k
k = 1.67 M min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
[A] = [A]₀ − kt
50 = 100 - (1.67)t
-50 = - 1.67t
t = 29.94 min
Answer:
0.499atm
Explanation:
The formula is
P1/V1 = P2/V2
so:
1.26atm/7.40L = P2/2.93L
then:
(1.26atm/7.40L)*2.93L = P2
= 0.4988918911atm
the answer must have 3 sig figs
125 Each half life it divides by 2 the amount
1000/2=500
500/2=250
250/2=125
Answer:
p3=0.36atm (partial pressure of NOCl)
Explanation:
2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) Kp = 51
lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively
![Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BNOCl%5D%5E%7B2%7D%20%7D%7B%5BNO%5D%5E%7B2%7D%20%5BCl_2%5D%20%7D)
![Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5Bp3%5D%5E%7B2%7D%20%7D%7B%5Bp1%5D%5E%7B2%7D%20%5Bp2%5D%20%7D)
p1=0.125atm;
p2=0.165atm;
p3=?
Kp=51;
On solving;
p3=0.36atm (partial pressure of NOCl)
