Chemistry, 03.01.2020 02:31, Emptypockets451

For the reaction x2 + y + z -> xy + xz it is found that doubling the concentration of x2 doubles the reaction rate. triplingthe concentration on triples the rate, and doubling the concentration of 2 has no effect, (a) what is the rate law for thisreaction? (b) why is it that the change in the concentration of 2 has no effect on the rate? (c) suggest a mechanism for thereaction that is consistent with the rate law.

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answered: bri1334

(a) R=k[X2][Y] (b) reaction is zero order wrt Z (c) X2 + Y --- XY + X (slow step)

X + Z --- XZ (fast step)

Explanation:

(a) Suppose the reaction rate R with respect to a component X2 with concentration [X2] is generally expressed as follows:

R = k [X2]^n (1)

Where k is the rate constant and n is the order of reaction with respect to X2.

When the reaction rate is found to double by doubling the concentration of X2, the following equation could be developed:

2R = k *(2[X2])^n (2)

Dividing equation (2) by (1) yields

2R/R= k *(2[X2])^n / k *[X2]^n

2=2^n , n = 1

Thus, the reaction is first order with respect to X2.

In the same manner,

Tripling the concentration of Y triples the rate,

R = k [Y]^m (3)

Where k is the rate constant and m is the order of reaction with respect to Y.

When the reaction rate is found to triple by tripling the concentration of Y, the following equation could be developed:

2R = k *(3[Y])^m (4)

Dividing equation (4) by (3) yields

3R/R= k *(3[Y])^m / k *[Y]^m

3=3^m , m = 1

Thus, the reaction is first order with respect to Y.

Therefore,the rate law for this reaction is R = k [X2] [Y]

(b)

Provided that doubling the concentration of Z has no effect, we perform similar analysis as above with p representing order of reaction wrt Z:

R = k [Z]^p (5)

R = k (2*[Z]^p (6)

R/R= k *(2[Z])^p / k *[Z]^p

1=2^p , p = 0

Thus, the reaction is zero order wrt Z. This explains why the change in concentration has no effect on the rate.

X2 + Y XY + X (slow step)

X + Z XZ (fast step)

The rate law is determined by the slow step, and this is very consistent with the experimentally observed data.

Besides, addition of the two step yields the overall reaction, and both steps are reasonable.

answered: Guest

A) the enthralling of formation for pure element in its standard state is always positivehope this : )

answered: Guest

The closer the atoms are to becoming ionic the more highly dense the negatively charged side gets.

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For the reaction x2 + y + z -> xy + xz it is found that doubling the concentration of x2 doubles...