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Likurg_2 [28]
3 years ago
15

A + B → C Select the rate law for the reaction above using the following information: Holding the concentration of A constant an

d doubling the concentration of B results in the rate of the reaction increasing from 1.5E-3 M/s to 1.2E-2 M/s. Keeping the concentration of B constant and doubling A results in the rate of the reaction increasing from 1.5E-3 M/s to 3.0E-3 M/s.
Chemistry
1 answer:
enyata [817]3 years ago
3 0
For the initial conditions:
1.5E-3 = k A^n B^m

For the second condition:
1.2E-2 = k A^n (2B)^m
1.2E-2 = 2^m k A^n B^m
From the initial condition:
1.2E-2 = 2^m (1.5E-3)
m = 3.32

For the third condition:
3.0E-3 = k (2A)^n B^m
3.0E-3 = 2^n k A^n B^m
3.0E-3 = 2^n (1.5E-3)
n = 1

Therefore, the rate law is:
r = k A B^3.32
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This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

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Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

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Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

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When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

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