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A reactant breaks down on a catalyst surface by a zero… | StemPy Lab
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A reactant breaks down on a catalyst surface by a zero order reaction. At a concentration of 0.22 mol/L the rate is 3.0 × 10⁻³ mol L⁻¹ s⁻¹. How long does the concentration take to fall from 0.52 mol/L to 0.34 mol/L?
Chemistry
Medium
High School (11–12)
Chemical Kinetics › Integrated Rate Equations and Half-Life
AP
A-Level
IB
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StemPy Lab
A reactant breaks down on a catalyst surface by a zero order reaction. At a concentration of
0.22
mol/L
0.22\ \text{mol/L}
0.22
mol/L
the rate is
3.0
×
10
−
3
mol L
−
1
s
−
1
3.0 \times 10^{-3}\ \text{mol L}^{-1}\,\text{s}^{-1}
3.0
×
1
0
−
3
mol L
−
1
s
−
1
. How long does the concentration take to fall from
0.52
mol/L
0.52\ \text{mol/L}
0.52
mol/L
to
0.34
mol/L
0.34\ \text{mol/L}
0.34
mol/L
?
173
s
173\ \text{s}
173
s
113
s
113\ \text{s}
113
s
60
s
60\ \text{s}
60
s
21
s
21\ \text{s}
21
s
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← Which of these statements are correct? (i) The unit of k for a first order reaction stays the same when the concentration is measured in bar instead of mol/L. (ii) The half-life of a zero order reaction doubles when the start concentration doubles. (iii) A first order reaction is about 94 % complete after four half-lives. (iv) For a zero order reaction a graph of ln[R] against time is a straight line.
In the reaction 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), ammonia is used up at 48 mmol L⁻¹ s⁻¹. At what rate is oxygen used up? →
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Chemical Kinetics › Integrated Rate Equations and Half-Life
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