Chemical Equilibrium: Quiz

Chemical Equilibrium: Quiz

7 questions

1. Question 1EasyWhich statement describes a system at chemical equilibrium?

   AThe reactions have stoppedBThe concentrations of reactants and products are equalCThe forward and reverse rates are equal, and concentrations are constantDAll the reactants have turned into products
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   Answer: The forward and reverse rates are equal, and concentrations are constant

   Equilibrium is dynamic: both reactions continue at the same rate, so concentrations stay constant, but they are generally not equal.

2. Question 2EasyWhat is $K_\text{c}$ for $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$?

   A$\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$B$\dfrac{2[\ce{SO3}]}{2[\ce{SO2}][\ce{O2}]}$C$\dfrac{[\ce{SO2}]^2[\ce{O2}]}{[\ce{SO3}]^2}$D$\dfrac{[\ce{SO3}]}{[\ce{SO2}][\ce{O2}]}$
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   Answer: $\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$

   Products over reactants, each raised to the power of its coefficient.

3. Question 3MediumWhat is $K_\text{c}$ for $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$?

   A$\dfrac{[\ce{CaO}][\ce{CO2}]}{[\ce{CaCO3}]}$B$[\ce{CO2}]$C$\dfrac{1}{[\ce{CO2}]}$D$[\ce{CaO}][\ce{CO2}]$
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   Answer: $[\ce{CO2}]$

   Pure solids are left out of the expression, so only the gas remains.

4. Question 4MediumAn equilibrium mixture of $\ce{N2O4 <=> 2NO2}$ contains 0.0450 M $\ce{N2O4}$ and 0.0300 M $\ce{NO2}$. What is $K_\text{c}$?

   A0.667B0.0200C1.33D50.0
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   Answer: 0.0200

   $K_\text{c} = (0.0300\ \text{M})^2 / (0.0450\ \text{M})$, with each concentration divided by 1 M: 0.000900 ÷ 0.0450 = 0.0200. The answer 0.667 forgets to square [NO₂].

5. Question 5MediumFor a reaction, $K_\text{c} = 49.0$ and $Q = 16$. What happens?

   AThe system is at equilibriumBThe reaction goes forward, making more productsCThe reaction goes in reverse, making more reactantsDK increases to 16
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   Answer: The reaction goes forward, making more products

   $Q < K$ means there are too few products, so the reaction goes forward until $Q = K$.

6. Question 6HardFor $\ce{2SO2 + O2 <=> 2SO3}$, $K_\text{c} = 800$. What is $K_\text{c}$ for $\ce{SO2 + 1/2 O2 <=> SO3}$?

   A400B28.3C1600D0.00125
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   Answer: 28.3

   Halving all the coefficients raises K to the power ½: √800 = 28.3. The answer 0.00125 is K for the reverse reaction.

7. Question 7Hard0.100 M $\ce{H2}$ and 0.100 M $\ce{I2}$ react: $\ce{H2 + I2 <=> 2HI}$, $K_\text{c} = 49.0$. What is [HI] at equilibrium?

   A0.0222 MB0.0778 MC0.156 MD0.200 M
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   Answer: 0.156 M

   Taking the square root of 49.0 = (2x)² / (0.100 M − x)² gives 7.00 = 2x / (0.100 M − x), so x = 0.0778 M and [HI] = 2x = 0.156 M. 0.0778 M is x itself.