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3 years ago

Consider the equation:

Chemistry

1 answer:

Dmitry [639]3 years ago

5 0

Answer:1. $Rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}$

2. The rate constant (k) for the reaction is $3.50M^\frac{-1}{2}s^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$rate=k[CHCl_3]^x[Cl_2]^y$

k= rate constant

x = order with respect to $CHCl_3$

y = order with respect to $Cl_2$

n = x+y= Total order

1. a) From trial 1: $0.0035=k[0.010]^x[0.010]^y$ (1)

From trial 2: $0.0069=k[0.020]^x[0.010]^y$ (2)

Dividing 2 by 1 : $\frac{0.0069}{0.035}=\frac{k[0.020]^x[0.010]^y}{k[0.010]^x[0.010]^y}$

$2=2^x,2^1=2^x$ therefore x=1.

b) From trial 2: $0.0069=k[0.020]^x[0.010]^y$ (3)

From trial 3: $0.0098=k[0.020]^x[0.020]^y$ (4)

Dividing 4 by 3: $\frac{0.0098}{0.0069}=\frac{k[0.020]^x[0.020]^y}{k[0.020]^x[0.010]^y}$

$1.4=2^y,2^{\frac{1}{2}}=2^y$ therefore $y=\frac{1}{2}$

$rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}$

2. to find rate constant using trial 1:

$0.0035=k[0.010]^1[0.010]^\frac{1}{2}$

$k=3.50M^\frac{-1}{2}s^{-1}$

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