2005 Exam 2

# 2005 Exam 2 - CHE132 Midterm 2 Form 0 CHE132 Midterm 2...

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CHE132 Midterm 2 Form 0 1 CHE132 Midterm 2 Spring 2005 Form 0 There are 20 questions. Each question is worth 5 points. The entire exam is worth 100 points. The first 18 questions are multiple choice. The last 2 questions require written answers. For full credit, show your work in the space provided for each of these questions on the answer form and write the answer in the box. Note: For question19, use the space provided on the FRONT of the answer form. For question 20, use the space provided on the BACK of the answer form. Constants The Gas constant = R = 8.314 J mol -1 K -1 The Gas constant = R = 0.0821 L atm mol -1 K -1 The molar mass of water is 18.02 g mol -1 The density of liquid water is 1 g/mL. 1 atm = 1.013 x 10 5 Pa. 1 cm 3 = 1 x10 -3 L = 1 x 10 -6 m 3 Equations For a first order reaction: ln([A] o /[A]) = kt For a second order reaction: 1/[A] – 1[A] o = kt reaction = coeff p f (products) - coeff r f (reactants) G° = -RTlnK eq lnK eq = - H°/RT + S°/R G reaction = reaction + RTlnQ k = Ae -Ea/RT lnk = lnA – (E a /RT) E R k k T T a ln 2 1 1 2 1 1 Quadratic Equation ax 2 + bx + c = 0 x b b ac a   2 4 2

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CHE132 Midterm 2 Form 0 2 Question 1. The rate law for a reaction is: Rate = k[A] 2 [B] If the concentration of A is doubled and the concentration of B is halved, what happens to the reaction rate? Which one of the following statements is correct? A: The reaction rate remains the same. B: The reaction rate is doubled. C: The reaction rate is halved. D: The reaction rate increases by 1.5 fold. E: The reaction rate decreases by 1.5 fold. B Question 2. If the initial concentration of the reactant in a first order reaction is 0.5 M and the half-life is 400 s, how long would it take for the concentration of the reactant to decrease to 0.062 M? A: 1205 s B: 400 s C: 4655 s D: 577 s E: .25 M A k = 0.693/400 = 1.7325 x 10 -3 s. ln(A 0 /A)=kt, t = (ln(0.5/0.062))/1.7325 x 10 -3 = 1205 s Question 3. The reaction of methyl bromide with hydroxide ion is first order in methyl bromide and first order in hydroxide ion: Rate = k[CH 3 Br][OH - ] CH 3 Br + OH - CH 3 OH + Br - In an isolation experiment, the initial concentrations of reactants are [CH 3 Br] = 3.7 x 10 -4 M and [OH - ] = 0.5 M. Under these conditions, the observed rate constant (k obs ) for the disappearance of CH 3 Br is 12 s -1 . Calculate the true rate constant (k) for the reaction. A: 32432 M -1 s -1 B: 6 M -1 s -1 C: 1.85 x 10 -4 M -1 s -1 D: 12 M -1 s -1 E: 24 M -1 s -1 E k obs = k[OH - ] so k = 12/0.5 = 24 M -1 s -1
CHE132 Midterm 2 Form 0 3 Question 4 The rate law for a reaction is: Rate = k[A] 0.5 [B] 2 Which one of the following statements is not correct? A: The reaction is 0.5 order in A.

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## This note was uploaded on 04/23/2010 for the course CHE 132 taught by Professor Hanson during the Spring '08 term at SUNY Stony Brook.

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2005 Exam 2 - CHE132 Midterm 2 Form 0 CHE132 Midterm 2...

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