The reaction C₂H₁ (9) + HI(g) →→ C₂H5I(g) is first order in both C₂H4 and HI, with t= k 1.03 x 10-5 at 328 °C. If 1.00 L of C₂H4 (9) at a concentration of 0.0120 M is rapidly mixed with the same volume of HI(g) also at a concentration of 0.0120 M, what is the time (in seconds) required for the C₂H4 concentration to decrease to a value of 0.00372 M? L mol s S

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The reaction
is first order in both
C₂H4 and
HI, with
t=
EM 189
C₂H₁ (g) + HI(g) → C₂H5I(g)
at 328 °C. If 1.00 L of
C₂H4 (9) at a concentration of 0.0120 M is rapidly mixed with the same volume of
HI(g) also at a concentration of 0.0120 M, what is the time (in seconds) required for the
C₂H4 concentration to decrease a value of 0.00372 M?
k=1.03 x 10-5
M-
T
Show Transcribed Text
S
First calculate the concentrations of the two reactants immediately after mixing. Since the volume of each gas is effectively doubled, the molarity will be cut in half:
0.00230 M
-0.00115 M
The two reactants start at the same concentration and react in a 1:1 ratio. Therefore at any point in the reaction,
T-SH). Let this concentration be
cand let the initial concentration be
Then the integrated rate law for the reaction is:
2.M
fo
L
mol-s
Substitute the values given into this equation
-1820M
-1.15 10 M
4-264×10²
THO
Incorrect
1
182 10 M 1.15 10 M
Solving for r gives t-1.75 -10% s
Ć
Transcribed Image Text:The reaction is first order in both C₂H4 and HI, with t= EM 189 C₂H₁ (g) + HI(g) → C₂H5I(g) at 328 °C. If 1.00 L of C₂H4 (9) at a concentration of 0.0120 M is rapidly mixed with the same volume of HI(g) also at a concentration of 0.0120 M, what is the time (in seconds) required for the C₂H4 concentration to decrease a value of 0.00372 M? k=1.03 x 10-5 M- T Show Transcribed Text S First calculate the concentrations of the two reactants immediately after mixing. Since the volume of each gas is effectively doubled, the molarity will be cut in half: 0.00230 M -0.00115 M The two reactants start at the same concentration and react in a 1:1 ratio. Therefore at any point in the reaction, T-SH). Let this concentration be cand let the initial concentration be Then the integrated rate law for the reaction is: 2.M fo L mol-s Substitute the values given into this equation -1820M -1.15 10 M 4-264×10² THO Incorrect 1 182 10 M 1.15 10 M Solving for r gives t-1.75 -10% s Ć
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