The reaction $CHCl_3(g) + Cl_2(g) \to CCl_4(g) + HCl(g)$ is first order in $CHCl_3$ and one-half order in $Cl_2$. The overall order is:
A1.5
B1
C2
D0.5
Answer & Solution
Correct answer: A. 1.5
Overall order = sum of orders = $1 + 1/2 = 3/2 = 1.5$. The rate law is rate = $k[CHCl_3][Cl_2]^{1/2}$.
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