Asked by sstt
2 NO(g) + Br2(g) 2 NOBr
A student performs three experiments studying initial concentrations and initial rates. The data is summarized in the table below.
[NO] [Br2] Rate (mol/Ls)
0.80 0.60 1.4 x 10-1
1.60 0.60 2.8 x 10-1
0.80 1.20 5.6 x 10-1
a) What is the order of this reaction with respect to NO? Explain how you know.
b) What is the order of this reaction with respect to Br2? Explain how you know.
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a) First order in NO.
Explanation: Compare experiments 1 and 2: [NO] doubles (0.80 → 1.60) while [Br2] is constant. Rate doubles (1.4×10^-1 → 2.8×10^-1), so rate ∝ [NO]^1.
b) Second order in Br2.
Explanation: Compare experiments 1 and 3: [Br2] doubles (0.60 → 1.20) while [NO] is constant. Rate increases by a factor of 4 (1.4×10^-1 → 5.6×10^-1), so rate ∝ [Br2]^2.
Thus the rate law is: rate = k[NO]^1[Br2]^2.
Explanation: Compare experiments 1 and 2: [NO] doubles (0.80 → 1.60) while [Br2] is constant. Rate doubles (1.4×10^-1 → 2.8×10^-1), so rate ∝ [NO]^1.
b) Second order in Br2.
Explanation: Compare experiments 1 and 3: [Br2] doubles (0.60 → 1.20) while [NO] is constant. Rate increases by a factor of 4 (1.4×10^-1 → 5.6×10^-1), so rate ∝ [Br2]^2.
Thus the rate law is: rate = k[NO]^1[Br2]^2.
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