Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Chemical Kinetics

Question:

For a reaction, 2A + 3B → C + 4D, the following expression log(-\(\frac{dA}{dt}\)) = log\(\frac{dD}{dt}\) + x is valid. What is the value of x?

Options:

0.3

-0.3

0.6

-0.6

Correct Answer:

-0.3

Explanation:

2A + 3B → C + 4D

rate = -\(\frac{1}{2}\)\(\frac{dA}{dt}\) = -\(\frac{1}{3}\)\(\frac{dB}{dt}\) = +\(\frac{1}{1}\)\(\frac{dC}{dt}\) = +\(\frac{1}{4}\)\(\frac{dD}{dt}\)

-\(\frac{1}{2}\)\(\frac{dA}{dt}\) = +\(\frac{1}{4}\)\(\frac{dD}{dt}\)

-\(\frac{dA}{dt}\) = \(\frac{1}{2}\)\(\frac{dD}{dt}\)

log(-\(\frac{dA}{dt}\)) = log(\(\frac{1}{2}\)\(\frac{dD}{dt}\))

log(-\(\frac{dA}{dt}\)) = log(\(\frac{dD}{dt}\)) - log2

log(-\(\frac{dA}{dt}\)) = log(\(\frac{dD}{dt}\)) - 0.3

comparing with log(-\(\frac{dA}{dt}\)) = log\(\frac{dD}{dt}\) + x 

x = -0.3