A certain reaction is known to have an activation energy E=38.0 kJ/mol. Furthermore, the rate of this reaction is measured at 265. K and found to be 3.2 x 10¹ 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 raised by 5% from 265. K to 278. K. How will the rate of the reaction change? Suppose the concentrations of all reactants is kept the same, but the catalyst is added, which has the effect of lowering the activation energy by 5%, from 38.0 kJ/mol to 36.1 kJ/mol. How will the rate of the reaction change? The rate will The rate will choose one choose one

The rate of a certain reaction is given by the following rate law: rate =k[N,]H, ° Use this information to answer the questions below. olo What is the reaction order in N,? x10 Ar What is the reaction order in H,? What is overall reaction order? At a certain concentration of N, and H2, the initial rate of reaction is 0.090 M / s. What would the initial rate of the reaction be if the concentration of N, were doubled? Round M S your answer to 2 significant digits. The rate of the reaction is measured to be 95.0 M / s when [N2] = 0.99 M and [H2] = 1.1 M. Calculate the value of the -2 -1 k = •S rate constant. Round your answer to 2 significant digits.

Some measurements of the initial rate of a certain reaction are given in the table below. H2 2 initial rate of reaction 1.10M| 2.49M 9.00 x 10°M/s 3.05M 2.49М 2.50 x 10*M/s 1.10M 6.51M 2.35 x 10*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|| x10 k = ?

O KINETICS AND EQUILIBRIUM Deducing a rate law from initial reaction rate data Some measurements of the initial rate of a certain reaction are given in the table below. H, I, initial rate of reaction 0.735 M 0.260 M 78.0M/s 0.735 M 0.341 M 102. M/s 0.354 M 0.260 M 37.6M/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. Do rate = k| x10 k = Explanation Check © 2021 McGraw-Hill Educat II

Draw the order of reaction graphs (reaction rate vs the concentration) for [NO] and [H2]. Explain the difference between the two graphs

16.  The decomposition of hydrogen bromide is 2 HBr(g)    →    H2(g)  +  Br2(g) The reaction has rate constants of 7.23 x 10-8 M-1 s-1 at 5.00 x 102 K and 9.14 x 10-6 M-1 s-1 at 6.00 x 102 K. What  is the activation energy in kJ/mol? Group of answer choices 136 kJ/mol none of the answers are correct 145 kJ/mol 107 kJ/mol 121 kJ/mol

A certain catalyzed reaction is known to have an activation energy E= 11.0 kJ/mol. Furthermore, the rate of this reaction is measured at 296. K and found to be 1.1 × 10 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 raised by 10% from 296. K to 326. 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 11.0 kJ/mol to 11.6 kJ/mol. How will the rate of the reaction change? The rate will The rate will choose one 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% V

The rate constant k for a certain reaction is measured at two different temperatures: temperature k 59.0 °C 6.5 x 101 14 163.0 °C |4.2 × 10 Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy E, for this reaction. a Round your answer to 2 significant digits. kJ E x10 mol

A certain catalyzed reaction is Known to have an activation energy Ea=29.0 kJ/mol. Furthermore, the rate of this reaction is measured at 331 K and found to be 2.7× 10* 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 raised by 10% from 331 K to 364 K. How will the rate or the reaction change? Suppose the concentrations of all reactants is kept the same, but the catalyst is removed which has the eftect of raising the activation energy by 10%, from 29.0 kJ/mol to 31.9 kJ mol How will the rate or the reaction change?