hous Lork Explain onsuer une alternative that best completes the statement or an
ID: 227742 • Letter: H
Question
hous Lork Explain onsuer une alternative that best completes the statement or answers the question. A) A B) A catalyst catalyst increases the rate of reaction, but is not consume increases the rate of the forward reaction, but does not alter the reverse rate C) A catalyst alters the mechanism of reaction. D) A catalyst alters the activation energy E) A catalyst may be altered in the reaction, but is always regencrated. 2) In which of the following solvents would you expect KBr to he most soluble? A) C6H14 hexane) B) CoH6 (benzene C) CH12 (cyclohexane D) CH,CH OH (ethanol) E) CCI4 (carbon tetrachloride) 3) 3) The activation energy for the reaction CHOO CH3 + CO is 71 kJ/mol. How many times greater is the rate constant for this reaction at 170°C than at 150 C? A) I.I B) 0.40 C) 5.0 D) 4.0 E) 2.5 4) At a certain temperature, the data below were collected for the reaction below A) Determine the rate law for the reaction. Initial concentrations (M ICI) 0.10 0.20 0.10 MolL-s 0.0015 0.0030 0.00075 0.10 0.10 0.050 A) None of the above B) rate klICI 11H21 C) rate kIICIIH2P D) rate kjICIP[H:12 E) rate = k[ ICI PlHIExplanation / Answer
1. The catalyst triggers the reaction rate by tuning the activation energy of the reactants to form the correspoinding products....(D) is the correct option.
2. To evaluate the solubility the thumb rule is that the polar solutes are soluble in polar solvents and vice versa. As KBr is polar it is only soluble in ethyl alcohol (polar solvent) and the correct option is (D).
3. According to Arrhenius equation
Ka=Aexp(-Ea/RT)
ln(Ka1/Ka2)=-Ea/R (1/T1-1/T2)
Ea= 17KJ/mole
ln(Ka1/Ka2)=-71000/8.314(1/443-1/423) = -(0.00225733-0.0023640666)=0.911505
Ka1/Ka2= 2.48806.
The correct answer is (D).
4.
k1=[ICL]^m. [H2]^n
Tis is the rate law of a reaction which is of m-th order with respect to ICL and n-th order with respect to H2.
0.0015 = (0.10)^m.(0.10)^n.......1
0.0030= (0.20)^m.(0.10)^n.......2
0.00075=(0.10)^m.(0.050)^n......3
dividing 1 and 2,
0.0015/0.0030=(0.10/0.20)^m
m=1
From 1 and 3
0.0015/0.00075=(0.10/0.05)^n
n=1
Thus the rate equation is
K =[ICl][H2]
option is (B).
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