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Test2FormASolutions - Chem 1036 Test 2 THIS IS FORM A [FORM...

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Chem 1036 Test 2 Feb. 29, 2008 THIS IS FORM A [FORM #s in ( )] Name _________________________ K w = 1.0 x 10 -14 Choose the single best answer to each question. 1.(5) The equilibrium constant, K c , for the decomposition of COBr 2 COBr 2 ( g ) CO( g ) + Br 2 ( g ) is 0.190. What is K c for the following reaction? 2CO( g ) + 2Br 2 ( g ) 2COBr 2 ( g ) 1. 0.0361 2. 2.63 3. 5.62 4. 10.5 5. 27.7 Answer: 5. 27.7 The second reaction is a result of reversing the first reaction and multiplying the coefficients by 2 . Reciprocal rule : reversing a reaction results in the new K value being the reciprocal of the first reaction: 1 K 5.263 0.190 == Coefficient rule: multiplying a reaction by a factor n, like 2, results in the new K value being equal to the first K value raised to the nth power. So raise K to the 2 power: K = (5.263) 2 = 27.7 Or 2 1 K 0.190 ⎛⎞ = ⎜⎟ ⎝⎠ = 27.7 2. (8) The equilibrium constant, K p , for the reaction H 2 ( g ) + I 2 ( g ) 2HI( g ) is 55.2 at 425°C. A rigid cylinder at that temperature contains 0.127 atm of hydrogen, 0.134 atm of iodine, and 1.055 atm of hydrogen iodide. Is the system at equilibrium? 1. Yes. 2. No, the forward reaction must proceed to establish equilibrium. 3. No, the reverse reaction must proceed to establish equilibrium. 4. Need to know the volume of the container before deciding.
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Answer: 3. No, the reverse reaction must proceed to establish equilibrium . You need to determine if the ratio of products to reactants is the proper ratio which is 55.2, the value of K. Calculate the reaction quotient , Q, to see how the ratio of initial concentrations compares to the ratio at equilibrium which is 55.2: 2 2 22 [HI] [1.055] Q = = = 65.4 [H ][I ] [0.127][0.134] Q = 65.4 K = 55.2 Q > K The ratio at this point is too large . The reaction will proceed in the direction necessary to decrease the product/reactant ratio. For the ratio to become smaller, the denominator (reactants) must increase as the numerator (the products) decreases. The reaction is proceeding to the left to decrease the amount of products and increase the amounts of reactants so the product/reactant ratio will decrease. 3.(6) Nitric oxide and bromine were allowed to react in a sealed container. When equilibrium was reached, the following partial pressures of the three gases were measured: NO: 0.526 atm; Br 2 : 1.59 atm; NOBr: 7.68 atm. Calculate K p for the reaction. 2NO( g ) + Br 2 ( g ) 2NOBr( g ) 1. 7.45 x 10 -3 2. 0.109 3. 9.18 4. 91.8 5. 134 Answer: 5. 134 2 2 2 [NOBr] K = [NO] [Br ] You have been given the equilibrium concentrations of all of the substances so the equilibrium expression can be solved for the value of K: 2 2 2 [NOBr] K = [NO] [Br ] 2 2 [7.68] K = [0.526] [1.59] = 134
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4.(9) A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br 2 ( g ) + I 2 ( g ) 2IBr( g ) When the mixture has come to equilibrium, the concentration of iodine monobromide is 1.190 M . What is the equilibrium constant for this reaction at 350°C? 1. 3.55 x 10 -3 2. 1.24 3. 1.47 4. 282 5. 325 Answer: 4. 282 2 22 [IBr] K = [Br ][I ] You need the equilibrium amount of reactants and products Set up a table to get equilibrium concentrations: You know the initial concentrations of IBr, Br 2 and I 2 and the equilibrium concentration of IBr: Br 2 ( g ) + I 2 ( g ) 2IBr( g ) I nitial 0.600 M 1.600 M 0 C hange ____________________________ _______ E qui 1.190 M Complete the table: Br 2 ( g ) + I 2 ( g ) 2IBr( g ) I nitial 0.600 M 1.600 M 0 C hange - 0.595M
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Test2FormASolutions - Chem 1036 Test 2 THIS IS FORM A [FORM...

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