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3 years ago

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The reaction A + 2B → products was found to follow the rate law: rate = k[A] 2[B]. Predict by what factor the rate of reaction w

ill increase when the concentration of A is doubled, the concentration of B is tripled, and the temperature remains constant

1 answer:

Alexus [3.1K]3 years ago

5 0

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of A and the of the concentration of B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

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<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

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pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

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$5.92=-\log[OH^-]$

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The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

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$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

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The expression of $K_b$ for above equation follows:

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Putting values in above expression, we get:

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To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

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Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

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$K_a\text{ for HF}=6.8\times 10^{-6}$

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So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

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