Question

An equilibrium mixture contains 0.710 mol HI, 0.460 mol I2, and 0.250 mol H2 in a 1.00-L flask.

What is the equilibrium constant for the following reaction? 2HI(g) H2(g) + I2(g)

K =

How many moles of I2 must be removed in order to double the number of moles of H2 at equilibrium?

_______ mol I2

Answer #1

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At 400 K, an equilibrium mixture of H2, I2, and HI consists of
0.082 mol H2, 0.084 mol I2, and 0.15 mol HI in a 2.50-L flask. What
is the value of Kp for the following equilibrium? (R = 0.0821 L ·
atm/(K · mol))
2HI(g) H2(g) + I2(g)
A. 0.045
B. 7.0
C. 22
D. 0.29
E. 3.4

0.250 moles of H2 and 0.250 moles of I2 were placed in a 1.00 L
flask at 500˚. The equilibrium constant, Kc, for the reaction
H2(g)+I2(g)-> 2HI(g) is 54.3. Calculate the equilibrium
concentrations of all species.

H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.71 L flask at a
certain temperature initially contains 0.760 g H2 and 96.8 g I2. At
equilibrium, the flask contains 90.5 g HI. Calculate the
equilibrium constant (Kc) for the reaction at this temperature.

H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.75 L flask at a
certain temperature initially contains 0.760 g H2 and 96.9 g I2. At
equilibrium, the flask contains 90.4 g HI. Calculate the
equilibrium constant (Kc) for the reaction at this temperature.

. If 2.00 mol of H2 and 1.00 mol of I2 come to equilibrium at
this temperature (458ºC), how many moles of HI will be in the final
mixture? (First, consider this question: Why is the volume of the
container not needed for this problem?)
H2(g) + I2(g) 2HI(g) Kc = 50.3 at 458ºC

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 0.325 M HI,
4.36×10-2 M H2 and 4.36×10-2 M I2. What will be the concentrations
of the three gases once equilibrium has been reestablished, if
2.27×10-2 mol of I2(g) is added to the flask?
[HI] = _____M
[H2] = ____M
[I2] = _____M

Consider the reaction: H2(g)+I2(g)⇌2HI(g)
A reaction mixture in a 3.64 −L flask at 500 K initially
contains 0.376 g H2 and 17.97 g I2. At equilibrium, the flask
contains 17.76 g HI.
Part A Calculate the equilibrium constant at this
temperature.
I keep getting 13413.06 and its not right I'm running out of tries,
please help me.

Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction
mixture in a 3.63 L flask at a certain temperature initially
contains 0.767 g H2 and 97.0 g I2. At equilibrium, the flask
contains 90.6 g HI. Calculate the equilibrium constant (Kc) for the
reaction at this temperature. Express your answer using two
significant figures.

A 1.00 L container holds 0.015 mol of H2 (g) , 0.015 mol of I2
(g), and 0.015 mol of HI (g) at 721 K. What are the
concentrations(pressures) of H2 (g), I2 (g), and HI (g) after the
system achieved a state of equilibrium? The value of Kc is 50.0 for
reaction: H2 (g) + I2 (g) 2HI (g)

At a certain temperature, the equilibrium constant, Kc, for this
reaction is 53.3.
H2(g) + I2(g) <----> 2HI(g)
Kc=53.3
At this temperature, 0.400 mol of H2 and 0.400 mol of I2 were
placed in a 1.00-L container to react. What concentration of HI is
present at equilibrium?

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