Asample of helium occupies 0.56 l at 988 mmhg and 25 degrees celsius at the sample is transferred to a 1.0 l flask at the same temperature what will the gas pressure be in the new flask
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Ответ:
(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.
(c) A suggested mechanism for the reaction that is consistent with the rate law will therefore be
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.