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!
The question is incomplete. Complete question is attached below.
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Correct Answer: <em>Option 1) 2-pentene</em>
Reason:
Following are the IUPAC rules for naming the compound
1) Select the
longest carbon chain. In present case longest carbon chain has 5 carbon atom. Hence, it is a pentane derivative.
2) In case of alkene,
replace 'e' of alkane by 'ene'3) Give
lowest number to function group. In present case, it is double bond.
Applying above rules, the IUPAC name of compound is
2-pentene