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General Chemistry Question List

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Kc for the reaction below at 25°C is 4.8 x 10-6. Calculate the
equilibrium concentration (mol/L) of Cl2(g) if the initial
concentration of ICl(g) is 1.33 mol/L. There is no I2 or Cl2
initially present.

2ICl(g) <-> I2(g) + Cl2(g)

a. 6.4 x 10-6
b. 343
c. 3.2 x 10-6
d. 2.9 x 10-3
e. 5.8 x 10-3

2. Consider the following reaction at equilibrium:
2NH3(g) <-> N2(g) + 3H2(g) &#916;H° = +92.4 kJ

Adding N2(g) to the system at equilibrium will __________.

a. decrease the concentration of H2(g) at equilibrium
b. remove all of the H2(g)
c. cause the reaction to shift to the right
d. increase the value of the equilibrium constant
e. decrease the concentration of NH3(g) at equilibrium

3. Consider the following reaction at equilibrium:

2CO2(g) <-> 2CO(g) + O2(g) (Traingle sign)H° = 514 kJ
Increasing the temperature will ________.

a. decrease the concentration of CO2(g)
b. shift the reaction to the right
c. increase the concentration of CO
d. increase the concentration of O2(g)
e. decrease the value of the equilibrium constant

3. The effect of a catalyst on a chemical reaction is to
_______________.

a. accelerate the forward reaction only
b. lower the energy of the transition state
c. make reactions more exothermic
d. increase the entropy change associated with a reaction
e. react with product, effectively removing it and shifting the
equilibrium to the right

4. Consider the equilibrium at 25°C:
1A(g) + 2B(g) <-> 3C(g) + D(L) Kc = 3.25 x 10-2
What is the value of Kp?

a. 19.453700
b. 80.521090
c. 0.136915
d. 0.000156
e. 0.032500

5. Consider the equilibrium at 25°C:
2A(g) + 2B(g)<-> 1C(g) + D(L) Kp = 2.65 x 10-4
What is the value of Kc?

a. 4030182.628
b. 3.881
c. 1586.224730
d. 0.002291
e. 2379.556

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Solution Preview

1. Kc for the reaction below at 25°C is 4.8 x 10-6. Calculate the equilibrium concentration (mol/L) of Cl2(g) if the initial concentration of ICl(g) is 1.33 mol/L. There is no I2 or Cl2 initially present.

2ICl(g) <-> I2(g) + Cl2(g)

Explanation:

At the beginning, [ICl] = 1.33M, [Cl2] = 0M, and [I2] = 0M

If 2x = moles of ICl that react. Therefore, x = moles of Cl2 and I2 that form, right?

At equilibrium, [ICl] will drop by some factor, 2x.
At equilibrium, [Cl2] will increase by x.
At equilibrium, [I2] will increase by x.

Therefore, at equilibrium,

Kc = [I2][Cl2]/[ICl]^2 = 4.8x10-6

4.8x10-6 = (x)(x)/(1.33-2x)^2

4.8x10-6 = x^2/(1.33-2x)^2

If we take the square root of both sides:

2.2x10-3 = x/1.33-2x

Simplifying, we get:

x = 2.9x10-3 - 4.4x10-3 x

x = 2.9x10-3

Therefore, "d. 2.9 x 10-3" is correct. ...

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This solution provides explanations to 3 chemistry questions.

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