1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.311 M HI, 4.18×10-2 M H2 and 4.18×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.85×10-2 mol of I2(g) is added to the flask?
2) The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K.
PCl5(g) ->PCl3(g) + Cl2(g)
An equilibrium mixture of the three gases in a 1.00 L flask at
500 K contains 0.258 M
PCl5,
5.57×10-2 M
PCl3 and
5.57×10-2 M
Cl2. What will be the concentrations of
the three gases once equilibrium has been reestablished, if
2.90×10-2 mol of
PCl3(g) is added to the flask?
3) The equilibrium constant, K, for the following reaction is 1.48×10-2 at 505 K.
PCl5(g) ->PCl3(g) + Cl2(g)
An equilibrium mixture of the three gases in a
18.9 L container at 505 K
contains 0.268 M PCl5,
6.30×10-2 M
PCl3 and
6.30×10-2 M
Cl2. What will be the concentrations of
the three gases once equilibrium has been reestablished, if the
equilibrium mixture is compressed at constant temperature to a
volume of 9.03 L?
4) The equilibrium constant, K, for the following reaction is 2.53×10-2 at 519 K.
PCl5(g) ->PCl3(g) + Cl2(g)
An equilibrium mixture of the three gases in a
9.51 L container at 519 K
contains 0.303 M PCl5,
8.76×10-2 M
PCl3 and
8.76×10-2 M
Cl2. What will be the concentrations of
the three gases once equilibrium has been reestablished, if the
volume of the container is increased to 16.2
L?
Given reaction: 2HI(g) ----> H2(g) + I2(g)
Before addition of I2 (moles) 0.311 4.18×10-2 4.18×10-2
I2 addition : 2.85×10-2
Initial (moles) 0.311 4.18x10-2 7.03x10-2
Change +2a -a -a
Equilibrium 0.311+2a 4.18x10-2-a 7.03x10-2-a
Since we are adding I2 the reaction will move backward
infavor of the formation of
HI and therefore the disappearance of a part of H2 and
I2.
(note: Since, the volume of the flask is one liter, the number moles = molarity)
Therefore, equilibrium concentration K = [H2][I2]/[HI]2
1.80×10-2=
(4.18x10-2-a)(7.03x10-2-a)/(0.311+2a)2
Therefore, after solving we the following quadratic equation;
0.99a2 - 0.1345a + 0.0012 = 0
Roots: a= 0.126, 0.0096 ( since, 0.126 is too large, so we can take a value as 0.0096 ~ 0.01 )
Therefore,
2HI(g) ---->
H2(g) +
I2(g)
0.311+2a
4.18x10-2-a
7.03x10-2-a
0.311+2x0.01
4.18x10-2-0.01
7.03x10-2-0.01
0.331
3.18x10-2
6.03x10-2
Hence, the concentration of [HI] = 0.331 M
[H2] = 3.18 x10-2 M
[I2] = 6.03 x
10-2 M
1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g)...
The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K: H2(g) + I2(g) ---------->2HI(g) 1) Calculate the equilibrium concentrations of reactants and product when 0.309 moles of H2 and 0.309 moles of I2 are introduced into a 1.00 L vessel at 698 K. [H2] = M? [I2] = M? [HI] = M? 2.The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K: PCl5(g)------->PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a...
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.316 M HI, 4.24×10-2 M H2 and 4.24×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.21×10-2 mol of I2(g) is added to the flask?
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.302 M HI, 4.05×10-2 M H2 and 4.05×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.203 mol of HI(g) is added to the flask?
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.329 M HI, 4.41×10-2 M H2 and 4.41×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.54×10-2 mol of H2(g) is added to the flask? [HI] = M [H2] = M [I2] = M
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.319 M HI, 4.27×10-2 M H2 & 4.27×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.224 mol of HI(g) is added to the flask? [HI] = M [H2] = M [I2] = M please help me!
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) ----> H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.306 M HI, 4.10×10-2 M H2 and 4.10×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.208 mol of HI(g) is added to the flask? [HI] = ______ M [H2] = ______ M [I2] = ______M
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) goes to H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.308 M HI, 4.14×10-2 M H2 and 4.14×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.67×10-2 mol of I2(g) is added to the flask? [HI] = __M [H2] = __M [I2] = __M
The equilibrium constant, K, for the following reaction is 3.40×10-2 at 527 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 10.5 L container at 527 K contains 0.271 M PCl5, 9.60×10-2 M PCl3 and 9.60×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 19.5 L? [PCl5] = M [PCl3] = M [Cl2] = M
The equilibrium constant, K, for the following reaction is 2.35×10-2 at 517 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 11.3 L container at 517 K contains 0.269 M PCl5, 7.94×10-2 M PCl3 and 7.94×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 5.27 L? [PCl5] = M [PCl3] = M [Cl2] =...
The equilibrium constant, K, for the following reaction is 3.16×10-2 at 525 K. PCl5(g) <---> PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 18.5 L container at 525 K contains 0.232 M PCl5, 8.57×10-2 M PCl3 and 8.57×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 8.42 L? [PCl5] = M [PCl3] = M [Cl2]...