I got x=0.0388 and the concentrations to be 0.0112 and 0.0776 yet the answer key
ID: 575492 • Letter: I
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I got x=0.0388 and the concentrations to be 0.0112 and 0.0776 yet the answer key in my book says it's wrong can someone please explain question 51 Find the equilibrium Concentraions of A and B for each value of a and b. Assume that the initial concentration of A in each case is 1.0 mol L- and that no B is present at the beginning of the reaction. b, a = 2; b = 2 50. Consider the reaction and associated equilibrium constant: aA(aq) + bB(aq) CC(aq) K = 5.0 -- ind the equilibrium concentrations of A, B, and C for each value of a, b, and c. Assume that the initial concentrations of A and B are each 1.0 mol L-' and that no product is present at the beginning of the reaction. c. a = 2; b = 1, c = 1 (set up equation for x; don't solve) 51. For the reaction, K-0.538 at 31 3 K. (0,05-c) 2+ N,04(g) 0 2 NO2(g) If a reaction vessel initially contains an N20, concentration of 0.0500 mol L- at 313 K, what are the equilibrium concentra- tions of N20, and NO2 at 313 K? 52. For the reaction, K 246 at 598 K If a reaction mixture initially contains a CO concentration of 0.1500 mol L-1 and a Cl2 concentration of 0.175 mol L-1 at 598 K, what are the equilibrium concentrations of CO, Cl2, and -COCI, at 598 K? 53. Consider the reaction: NiO(s) + CO(g)-Ni(s) + CO2(g) K = 80.3 at 1500 K If a mixture of solid nickel(II) oxide and 2.28 bar carbon mon- oxide is allowed to come to eauilibrium at 150oKExplanation / Answer
51)
N2O4 (g) <------------> 2 NO2 (g)
0.0500 0
0.050 - x 2x
K = [NO2]^2 / [N2O4]
0.538 = (2x)^2 / 0.050 - x
0.0269 - 0.538 x = 4x^2
4x^2 + 0.538 x - 0.0269 = 0
x = 0.0388
concentrations :
[N2O4] = 0.0112 M
[NO2] = 0.0776 M
Note : your answers are 100 % correct. if you want to check put these values in equilibrium constant expression , then we get 0.538 . so these are correct.
K = (0.0776)^2 / 0.0112 = 0.538
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