Identify the reaction order from each of the following rate constants k = 2 . 3 × 10 - 5 L mol - 1 s - 1 [Hint. The order of a reaction can be identified from the units given for each value of k.]

The order of the reaction can be determined by the unit of rate constant, which is given by: k = mol L 1 - n time - 1 where n is the value of order of reaction. For the second-order reaction n = 2 , So k = mol L 1 - 2 time - 1 k = mol L - 1 time - 1 k = mol - 1 L time - 1 Thus, given in the question that k = 2 . 3 × 10 - 5 L mol - 1 s - 1 Here, the rate constant k is expressed with respect to mol - 1 L s - 1 So, it is a second-order reaction.

Identify the reaction order from each of the following rate constants k = 2 . 3 × 10 - 5 L mol - 1 s - 1 [Hint. The order of a reaction can be identified from the units given for each value of k.]

The order of the reaction can be determined by the unit of rate constant, which is given by: k = mol L 1 - n time - 1 where n is the value of order of reaction. For the second-order reaction n = 2 , So k = mol L 1 - 2 time - 1 k = mol L - 1 time - 1 k = mol - 1 L time - 1 Thus, given in the question that k = 2 . 3 × 10 - 5 L mol - 1 s - 1 Here, the rate constant k is expressed with respect to mol - 1 L s - 1 So, it is a second-order reaction.

Identify the reaction order from each of the following rate constants 3 × 10 - 3 mol L - 1 s - 1 [Hint. The order of a reaction can be identified from the units given for each value of k.]

The order of the reaction can be determined by the unit of rate constant, which is given by: k = mol L 1 - n time - 1 where n is the value of order of reaction. For zero-order reaction, n = 0 ∴ k = mol L 1 - 0 time - 1 k = mol L 1 time - 1 k = mol L - 1 time - 1 Thus, given in the question that k = 3 × 10 - 3 mol L - 1 s - 1 Here, the rate constant k is expressed with respect to mol L - 1 s - 1 . Thus, it is a zero-order reaction.

For the reaction, X + Y → Z , the rate is given as k [ X ] 1 / 3 [ Y ] 1 . Calculate the order of the reaction.

The reaction is: X + Y → Z By the rate law, the rate constant is given by k X m Y n . Here, given in the question, m = 1 3 and n = 1 Thus, order of the reaction = m + n = 1 3 + 1 = 4 3 Hence, it is a reaction of fractional order.

The reaction N 2 O 5 ⟶ 2 NO 2 + 1 2 O 2 is of first order in N 2 O 5 . Its rate constant is 6 . 2 × 10 - 6 s - . If in the beginning, N 2 O 5 is 15 mol L - , calculate the rate of reaction in the beginning.

The reaction N 2 O 5 ⟶ 2 NO 2 + 1 2 O 2 is of the first order i.e. n = 1 . Also, N 2 O 5 = 15 mol L - 1 in the beginning. So, the rate law expression by chemical reaction, we have, = k N 2 O 5 1 mol L - 1 Here, k = 6 . 2 × 10 - 6 s - (Given) By substituting the value of k and N 2 O 5 in the above relation, we get, the rate of reaction in the beginning = 6 . 2 × 10 - 6 s - 15 mol L - 1 = 9 . 3 × 10 - 5 mol L - 1 s - 1 Hence, the answer.

A first order reaction is found to have a rate constant k = 5 . 5 × 10 - 4 s - 1 . Find the half-life of the reaction.

The half-life of the first order reaction is given by: t 0 . 5 = 0 . 693 k Where the value of k = 5 . 5 × 10 - 4 s - (Given) Substitute the value of k in the above relation, t 0 . 5 = 0 . 693 5 . 5 × 10 - 4 s - = 0 . 693 × 10 4 5 . 5 s = 69300 55 = 1260 t 0 . 5 = 1 . 26 × 10 3 s Hence, the answer.