Question

Consider the reaction C + D products. The rate of the reaction increases by a factor of 4 when concentration of C is doubled. The rate of reaction is tripled when concentration of D is tripled. What is the order of the reaction? Write the rate law.

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

The order of reaction is 3.

Explanation:

cC + dD → Products

The rate law for the above given equation is:

Rate(R) = k $[C]^{c}$ $[D]^{d}$  ......................................... (1)

Here the order of the above given reaction is (c+d).

According to the first condition, the rate of the reaction increases by a factor of 4 when concentration of C is doubled. So,

4*R = k $[2C]^{c}$$[D]^{d}$  ................................ (2)

Now, dividing equation (2) by equation (1), we get:

$\frac{4R}{R}$ = $\frac{k[2C]^{c} [D]^{d}} {k[C]^{c} [D]^{d}}$

4 = $2^{c}$  ⇒  $2^{2}$  =  $2^{c}$  ⇒ c = 2

According to the second condition, the rate of reaction is tripled when concentration of D is tripled. So,

3*R = k $[C]^{c} [3D]^{d}$   .......................... (3)

Now, dividing equation (3) by equation (1), we get:

$\frac{3R}{R}$ =  $\frac{k[C]^{c}[3D]^{d}} {k[C]^{c}[D]^{d}}$  ⇒  3 = $3^{d}$   ⇒   $3^{1}$  =  $3^{d}$    ⇒  d  = 1

Thus, the order of the reaction is (c + d) which is equal to 3.

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