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Chapter 1: Problem 6

$$A+B \rightarrow C+D \quad \text { rate }=k[A][B]^{2}$$ What are the potential units for the rate constant for the above reaction? (A) \(\mathrm{s}^{-1}\) (B) \(\mathrm{s}^{-1} M^{-1}\) (C) \(\mathrm{s}^{-1} M^{-2}\) (D) \(\mathrm{s}^{-1} M^{-3}\)

Short Answer

Expert verified
(C) \(s^{-1}M^{-2}\)

Step by step solution

01

Understanding the Problem

The reaction given in the exercise is \(A + B \rightarrow C + D\) with a rate expression \( \text{rate} = k[A][B]^2\). The 'k' in the expression is the rate constant for the reaction which we need to find the units of. The order of the reaction is the sum of the powers of the concentrations of the reactants in the rate equation, which in this case is \(1 + 2 = 3\).
02

Calculating Units of the Rate Constant

The units of rate of reaction is \(M \cdot s^{-1}\) (Molarity per second). As the reaction is third order, the units of rate constant (k) will be calculated by factoring in the units of [A] and [B]^2. Since [A] has units of M (Molarity) in the denominator to balance the overall units and [B]^2 would have \(M^2\) (Molarity squared), the calculated unit of the rate constant is \(s^{-1}M^{-2}\).

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Most popular questions from this chapter

\(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{PCl}_{5}(g) \Delta H=-92.5 \mathrm{kJ} / \mathrm{mol}\) In which of the following ways could the reaction above be manipulated to create more product? (A) Decreasing the concentration of \(\mathrm{PCl}_{3}\) (B) Increasing the pressure (C) Increasing the temperature (D) None of the above

After 44 minutes, a sample of \(_{19}^{44} \mathrm{K}\) is found to have decayed to 25 percent of the original amount present. What is the half-life of \(_{19}^{44} \mathrm{K}?\) (A) 11 minutes (B) 22 minutes (C) 44 minutes (D) 66 minutes

$$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ The above reaction will experience a rate increase by the addition of a catalyst such as platinum. Which of the following best explains why? (A) The catalyst causes the value for \(\Delta G\) to become more negative. (B) The catalyst increases the percentage of collisions that occur at the proper orientation in the reactant molecules. (C) The catalyst introduces a new reaction mechanism for the reaction. (D) The catalyst increases the activation energy for the reaction.

Silver sulfate, \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) , has a solubility product constant of \(1.0 \times 10^{-5} .\) The below diagram shows the products of a precipitation reaction in which some silver sulfate was formed. (Diagram Can't Copy) Which ion concentrations below would have led the precipitate to form? (A) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (B) \(\left[\mathrm{Ag}^{+}\right]=0.10 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (C) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.10 M\) (D) This is impossible to determine without knowing the total volume of the solution.

Use the following information to answer questions 12-15. When heated in a closed container in the presence of a catalyst, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) If 12.25 g of potassium chlorate decomposes, how many grams of oxygen gas will be generated? (A) 1.60 g (B) 3.20 g (C) 4.80 g (D) 18.37 g
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