The rate law for a certain enzymatic reaction is zero order with respect to the substrate. The rate constant for this reaction is 4.1 x 104 M-s1. If the initial concentration of the substrate is 0.631 mol L1, what is the initial rate of the reaction? 2.5 rate = x 10 M.s-1

Refer to the following reaction and rate law equation. A + 2B →C r = k [A] [B]² What will happen to the overall rate of reaction if the concentration of A is doubled and the concentration of B is halved? The rate of reaction will be half the initial rate. The rate of reaction will be quadruple (4x) the initial rate. The rate of reaction will remain the constant. The rate of reaction will be double (2x) the initial rate.. The rate of reaction will be a quarter of the initial rate.

Some measurements of the initial rate of a certain reaction are given in the table below. N2 H2 initial rate of reaction 2.22M 0.881M /93.0Ms 2.22M 2.90M ×1.01/103Ms 0.522M 0.881M /5.14Ms Use this information to write a rate law for this reaction, and calculate the value of the rate constant k . Round your value for the rate constant to 2 significant digits. Also be sure your answer has the correct unit symbol.   =ratek   =k

Some measurements of the initial rate of a certain reaction are given in the table below. N2 H2 initial rate of reaction 2.31M 0.547M 89.0M/s 11.0M 0.547M 424. M/s 2.31 M 1.75М 911. M/s Use this information to write a rate law for this reaction, and calculate the value of the rate constant k. Round your value for the rate constant to 3 significant digits. Also be sure your answer has the correct unit symbol. rate = k |I| x10 k = ?

Some measurements of the initial rate of a certain reaction are given in the table below. [H,] I initial rate of reaction 0.283 M 1.65 M 0.858 M/s 0.283 M 2.21 M 1.15 M/s 0.125 M 1.65 M 0.379 M/s Use this information to write a rate law for this reaction, and calculate the value of the rate constant k. Round your value for the rate constant to 3 significant digits. Also be sure your answer has the correct unit symbol.

A certain catalyzed reaction is known to have an activation energy E =39.0 kJ/mol. Furthermore, the rate of this reaction is measured at 278. K and found to be 4.7 × 10² M/s. Use this information to answer the questions in the table below. Suppose the concentrations of all reactants is kept the same, but the temperature is lowered by 10% from 278. K to 250. K. How will the rate of the reaction change? Suppose the concentrations of all reactants is kept the same, but the catalyst is removed, which has the effect of raising the activation energy by 5%, from 39.0 kJ/mol to 41.0 kJ/mol. How will the rate of the reaction change? The rate will The rate will choose one ✓ choose one stay the same rise about 5% rise more than 5% rise less than 5% fall about 5% fall more than 5% fall less than 5%

Only typed solution.

Some measurements of the initial rate of a certain reaction are given in the table below. [N₂] [H₂] initial rate of reaction 1.30M 0.612M 22.0M/s 3.84M 0.612M 192.M/s 1.30M 0.118M 0.818M/s Use this information to write a rate law for this reaction, and calculate the value of the rate constant k. Round your value for the rate constant to 3 significant digits. Also be sure your answer has the correct unit symbol. rate = k k= 0 ☐ x10 0 X

Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation 2H,O,(aq) 2 H₂O(1) + O₂(g) The activation energy for this reaction is 75 kJ.mol-¹. In the presence of a metal catalyst, the activation energy is lowered to 49 kJ.mol-¹. At what temperature would the non-catalyzed reaction need to be run to have a rate equal to that of the metal-catalyzed reaction at 25 °C? T = K

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E=39.0 kJ/mol. If the rate constant of this -1 reaction is 6.0 x 102 M's -1 at 9.0 °C, what will the rate constant be at 53.0 °C? Round your answer to 2 significant digits. -1 k = x10