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A gas X breaks down with the rate law rate = k[X]^(3/2)… | StemPy Lab
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A gas X breaks down with the rate law rate = k[X]^(3/2) in a rigid vessel. More X is pumped in until its pressure, and so its concentration, is four times the old value. How many times faster does the reaction run?
Chemistry
Medium
High School (11–12)
Chemical Kinetics › Rate of Reaction, Rate Law and Order
Board Exam
AP
A-Level
IB
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StemPy Lab
A gas X breaks down with the rate law
rate
=
k
[
X
]
3
/
2
\text{rate} = k[\text{X}]^{3/2}
rate
=
k
[
X
]
3/2
in a rigid vessel. More X is pumped in until its pressure, and so its concentration, is four times the old value. How many times faster does the reaction run?
16
16
16
8
8
8
6
6
6
4
4
4
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← At the moment when [A] = 0.35 mol/L, a first order reaction runs at 7.0 × 10⁻³ mol L⁻¹ s⁻¹. Find the rate constant.
Assertion (A): The order of a reaction can always be found from the coefficients in its balanced equation. Reason (R): The order of a reaction is found by experiment. →
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Chemical Kinetics › Rate of Reaction, Rate Law and Order
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