Chemical Kinetics Question 159
Question: For the reaction, $ 3A+2B\to C+D $ , the differential rate law can be written as:
Options:
A) $ \frac{1}{3}\frac{d[A]}{dt}=\frac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}} $
B) $ -\frac{d[A]}{dt}=\frac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}} $
C) $ +\frac{1}{3}\frac{d[A]}{dt}=-\frac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}} $
D) $ -\frac{1}{3}\frac{d[A]}{dt}=\frac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}} $
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Answer:
Correct Answer: D
Solution:
[d] For the reaction $ 3A+2B\xrightarrow{{}}C+D $ Rate of disappearance of A = Rate of appearance of C reaction $ =-\frac{1}{3}\frac{d[A]}{dt}=\frac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}} $
BETA
