(a)
Interpretation:
Which of the given options will change the value of rate constant of a given reaction has to be explained.
Concept introduction:
Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.
Rate law: It is an equation that related to the
- Depends on order of the
chemical reaction , the rate law or rate equation also varies.
To explain how will change the value of k when the concentration of
(b)
Interpretation:
Which of the given options will change the value of rate constant of a given reaction has to be explained.
Concept introduction:
Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.
Rate law: It is an equation that related to the rate of reaction to the concentrationsor pressures of substrates (reactants). It is also said to be as rate equation.
- Depends on order of the chemical reaction, the rate law or rate equation also varies.
To explain how will change the value of k when the reaction is run in an organic solvent
(c)
Interpretation:
Which of the given options will change the value of rate constant of a given reaction has to be explained.
Concept introduction:
Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.
Rate law: It is an equation that related to the rate of reaction to the concentrationsor pressures of substrates (reactants). It is also said to be as rate equation.
- Depends on order of the chemical reaction, the rate law or rate equation also varies.
To explain how will change the value of k when the volume of container is doubled
(d)
Interpretation:
Which of the given options will change the value of rate constant of a given reaction has to be explained.
Concept introduction:
Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.
Rate law: It is an equation that related to the rate of reaction to the concentrationsor pressures of substrates (reactants). It is also said to be as rate equation.
- Depends on order of the chemical reaction, the rate law or rate equation also varies.
To explain how will change the value of k when the temperature is decreased
(e)
Interpretation:
Which of the given options will change the value of rate constant of a given reaction has to be explained.
Concept introduction:
Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.
Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants). It is also said to be as rate equation.
- Depends on order of the chemical reaction, the rate law or rate equation also varies.
To explain how will change the value of k when a catalyst is added
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Chemistry: Atoms First V1
- The following equation represents a reversible decomposition: CaCO3(s)CaO(s)+CO2(g) Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3 remains?arrow_forwardConsider the following reaction: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) (a) The rate law for this reaction is second order in NO(g) and first order in H2(g). What is the rate law for this reaction?(b) If the rate constant for this reaction at a certain temperature is 79200, what is the reaction rate when [NO(g)] = 0.0852 M and [H2(g)] = 0.137 M?Rate =____ M/s.(c) What is the reaction rate when the concentration of NO(g) is doubled, to 0.170 M while the concentration of H2(g) is 0.137 M?Rate = ____ M/sarrow_forwardWhen the rate of the reversible reaction A + BSC is studied under a certain set of conditions, it is found that the rate of the forward reaction is kA]. What can be concluded about the rate law for the reverse reaction under these conditions? (A) Rate = k-[C] %3D (B) Rate = k [B] (C) The rate law of the reverse reaction cannot be determined from the information given. (D) An crror must have been made, since if the reaction is reversible, the forward rate law must be Rate = k{A][B]. %3Darrow_forward
- The decomposition reaction on N2O5 in carbon tetrachloride is 2 N2 O5-----> 4NO2 + O2. The rate law is first order in N2O5. At 64 °C the rate constant is 4.82 X 10-3s-1 (a) write the rate law for the reaction. (b) what is the rate when the concentration of N2O5 is doubled from 0.0240 M to 0.0480 M? Group of answer choices A) (a) rate= k[N2O5] (b) the rate doubles to 1.16 x 10E(-4) M/sec B.) (a) rate= k[N2O5] (b) the rate is halved to 5.8 x 10E(-5) M/sec C.) (a) rate= k[N2O5] (b) the rate doubles to 2.32 x 10E(-4) M/sec D.) (a) rate= k[N2O5]0 (b) the rate stays the samearrow_forwardConsider the following reaction: 4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g)(a) The rate law for this reaction is first order in HBr(g) and first order in O2(g). What is the rate law for this reaction?(b) If the rate constant for this reaction at a certain temperature is 8.80e+03, what is the reaction rate when [HBr(g)] = 0.00429 M and [O2(g)] = 0.00758 M?Rate = _______ M/s.(c) What is the reaction rate when the concentration of HBr(g) is doubled, to 0.00858 M while the concentration of O2(g) is 0.00758 M?Rate = _______ M/sarrow_forwardConsider the following reaction: 1. 2 N,O5 (g) → 4 NO, (g) + O, (g) The initial concentration of N2O5 was 0.48 mol/L, and 25 minutes after initiating the reaction, all of the N,Os has been consumed. (a) Calculate the average rate of the reaction over this 25-minute time interval. (b) Is it correct to assume that the rate law is Rate = k[N,O5]² based on the balanced chemical equation? Briefly explain your answer.arrow_forward
- The decomposition of phosphine, a very toxic gas, forms phosphorus and hydrogen in the following reaction: 4PH3(g) -> P4(g)+ 6H2(g) (a) Express the rate of the reaction with respect to each of the reactants and products.(b) if the instantaneous rate of the reaction with respect to PH3 is 0.34 M•S^-1 ,what is the instantaneous rate of the reaction?arrow_forwardConsider the reaction A + B ¡ C + D. Is each of the following statements true or false? (a) The rate law for the reaction must be Rate = k3A43B4. (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction.arrow_forwardAssume that the formation of nitrogen dioxide, 2 NO(g) + 02(g) – 2 NO2(g) is an elementary reaction. (a) Write the rate law for this reaction. (Rate expressions take the general form: rate = k. [A]ª . [B]b.) chemPad О Help Greek - rate=k•[NO]2.[02] rate=k*[NO]^2*[O_2] Correct. (b) A sample of air at a certain temperature is contaminated with 1.9 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. If not, repeat your answer from above. (Rate expressions take the general form: rate = k . [A]ª . [B]b. Use k' for the new rate constant as needed.) chemPad O Help Greek - rate=k':[NO]2 rate=k*[NO]^2 Correct. (c) Under the conditions described in (b), the half-life of the reaction has been estimated to be 6.7x103 min. What would the half-life be if the initial concentration of NO were 12.4 ppm? 4.0 |1030192 X min Supporting Materials Periodic Table Constants and E Supplemental Dataarrow_forward
- The following data was obtained from the reaction: (a) Determine the rate law. (b) Calculate the rate constant. (c) Calculate the initial rate, if [NO2~] = 0.1 M and [NH4+] = 0.1 Marrow_forwardConsider the following reaction: O2(g) + 2 NO(g) 2 NO2(g)(a) The rate law for this reaction is first order in O2(g) and second order in NO(g). What is the rate law for this reaction?(b) If the rate constant for this reaction at a certain temperature is 7840, what is the reaction rate when [O2(g)] = 0.0162 M and [NO(g)] = 0.0299 M?Rate = _____ M/s.(c) What is the reaction rate when the concentration of O2(g) is doubled, to 0.0324 M while the concentration of NO(g) is 0.0299 M?Rate = _____ M/sarrow_forwardFor a chemical reaction, mA → xB, the rate law is r = k[A]². If the concentration of A is doubled, the reaction rate will be (a) doubled (b) quadrupled (c) increased by 8 times (d) unchanged.arrow_forward
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