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3 years ago

Kinetic Energy Problems = KE = 0.5m x 12

Chemistry

1 answer:

Brut [27]3 years ago

8 0

Explanation:

The kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2$

1. Mass, m = 5 kg

Speed, v = 2 m/s

$K_1=\dfrac{1}{2}\times 5\times (2)^2\\\\=10\ J$

2. Mass, m = 71 kg

Speed, v = 1 m/s

$K_2=\dfrac{1}{2}\times 71\times (1)^2\\\\=35.5\ J$

3. Mass, m = 71 kg

Speed, v = 5 m/s

$K_3=\dfrac{1}{2}\times 71\times (5)^2\\\\=887.5\ J$

4. Mass, m = 1816 kg

Speed, v = 25 m/s

$K_4=\dfrac{1}{2}\times 1816\times (25)^2\\\\=567500\ J$

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Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la

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rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

$\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}$

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

$-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]$

2. Derive the rate law

Assume k₋₁ ≫ k₂.

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is

rate = k[A][B] where k = k₂K

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