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kotykmax [81]
3 years ago
8

Which statement is true regarding a catalyst?

Chemistry
1 answer:
Alla [95]3 years ago
4 0

Answer:

d). A catalyst lowers the activation energy for a chemical reaction.

Explanation:

From the given choices, it is true that a catalyst lowers the activation energy for a chemical reaction.

The activation energy is the energy barrier that must be overcome before a chemical reaction can occur.

Some reactions have very high activation energy and would not occur without the introduction of a catalyst.

The catalyst brings the reactants into contact by removing the energy deficiency in the system.

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2 HI(g) = H2(g) + 1,(9) K, - Party met = 0.0016 The decomposition of HI(g) at 298 K is represented by the equilibrium equation a
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Answer:

(C) P = 7.4 torr, because it is directly proportional to the initial pressure of HI.

Explanation:

From the equation of reaction,

Kc = [H2][I2]/[HI]^2

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Initial pressure of HI = 200 torr

Let the equilibrium pressure of I2 be y

Mole ratio of H2 to I2 from the equation of reaction is 1:1, equilibrium pressure of H2 is also y

Mole ratio of HI to I2 is 2:1, equilibrium pressure of HI is (200 - 2y)

0.0016 = y×y/(200 - 2y)^2

y^2 = 0.0016(40,000 - 800y + 4y^2)

y^2 = 64 - 1.28y + 0.0064y^2

y^2 - 0.0064y^2 + 1.28y - 64 = 0

0.9936y^2 + 1.28y - 64 = 0

y^2 + 1.29y - 64.41 = 0

The value of y must be positive and is obtained by the use of the quadratic formula.

y = [-1.29 + sqrt(1.29^2 - 4×1×-64.41)] ÷ 2(1) = [-1.29 + 16.10] ÷ 2 = 14.81 ÷ 2 = 7.4

Equilibrium pressure of I2 is 7.4 torr because the relationship between I2 and HI is direct in which increase in one quantity (initial pressure of HI) would result to a corresponding increase in the other quantity (equilibrium pressure of I2)

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