Answer:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
Explanation:
For the reaction:
H₂C₂O₄(g) → CO₂(g) + HCOOH(g)
At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:
H₂C₂O₄(g) = P₀ - x
CO₂(g) = x
HCOOH(g) = x
P at t=20000 is:
P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x
For 1st point:
x = 92,8-65,8 = 27
Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8
2nd point:
x = 130-92,1 = 37,9
H₂C₂O₄(g): 92,1 - 37,9 = 54,2
3rd point:
x = 157-111 = 46
H₂C₂O₄(g): 111-46 = 65
Now, as the rate law is :
v = k P[H₂C₂O₄]
Based on integrated rate law, k is:
(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k
1st point:
k = 2,64x10⁻⁵
2nd point:
k = 2,65x10⁻⁵
3rd point:
k = 2,68x10⁻⁵
The averrage of this values is:
k = 2,66x10⁻⁵
That means law is:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
I hope it helps!
No pressure
Ahhhhhh (filling character req)
<span> C.The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, so the theory had to be rejected. </span>
Insulation wraps because independent is the variable you are changing to affect the dependent variable (what you are measuring)
