Question

the following reaction 2NO(0) + 2H2 (g) № (g) + 2 HaO (g) H2/0000 M quilbrium constant, Ke, for the reaction. Initially, a mixture of 0.300 M NO, 0.100 M NOwas found to beg,062-м. D.ig2 the value of the brium (init h N2)t eubrhum the conentreton dr H20 was allowed to reach equilibrium ( 0-200 8) Consider the following reaction. HSO.. (aq) + H2O (I) → HO. (aq) + SO2-(aq) The initial concentrations are [HSO門-0.3000 M, [HO+]s 0.050 M. [SO (a) Which way would the reaction shift to reach equilibrium? (b) What are the equilibrium concentrations of the products and reactants. = 0.100 M.

Answer 1

7) Given reaction is:

2NO(g) + 2H₂(g) ---> N₂(g) + 2H₂O(g)

At equilibrium, [NO] can't be 0.162M. But it can be 0.062 M

Because if [NO] = 0.162 M and can't be greater [H₂]

	2NO+	2H2(g)	<=>	N2	2H2O
I(M)	0.300	0.100		0	0.200
C(M)	-2x = 0.062 x = 0.031	-2x		+x	+2x
E(M)	0.3 - 0.062 = 0.238	0.1-0.062 = 0.038		0.031	0.062

K_eq = [N₂][H₂O]² / [NO]²[H₂]² = (0.031)(0.062)² / (0.238)²(0.038)² = 1.46

2. Given reaction is:

HSO₄^-(aq) + H₂O(l) ---> H₃O⁺(aq) + SO₄^2-(aq)

[HSO₄^-] = 0.30 M

[H₃O⁺] = 0.050 M

[SO₄^2-] = 0.100 M

(a)For given reaction

Q_c = [SO₄^2-][H₃O⁺] /[HSO₄^-] = (0.100*0.050) / 0.30 = 0.017 > K

.So reaction will shift backward i.e. left to form reactant i.e. Q becomes equal to K

(b)

	[HSO4-]	H2O	<=>	H3O+	SO42-
I(M)	0.30			0.050	0.100
C(M)	+x			-x	-x
E(M)	0.30+x			0.050-x	0.100-x

K = [SO₄^2-][H₃O⁺] /[HSO₄^-] = (0.050-x)(0.100-x) / (0.30+x)

0.012(0.30+x) = 0.005 -0.15x + x²

0.0036 + 0.012 x = 0.005 -0.15x + x²

x² - 0.162 x + 0.0014 = 0

x = 0.00916 and 0.153 ( not possible, as x can't be greater than 0.05)

So, at equilibrium

[HSO₄^-] = 0.30 M + 0.00916 = 0.309 M

[H₃O⁺] = 0.050 M - 0.00916 = 0.0408 M

[SO₄^2-] = 0.100 M- 0.00916 = 0.0908 M