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question 4 (1 point) which is true of the numerical value of the equili…

Question

question 4 (1 point)
which is true of the numerical value of the equilibrium constant, $k_{eq}$?

  • it represents the relative concentration of products compared with reactants.
  • it represents the ratio of reactants to products.
  • it is measured in the si unit, mole.
  • both a and c are true.
  • both b and c are true.

question 5 (1 point)
which represents $k_{eq}$ for the equilibrium of the system shown below?
$\ce{n_{2}(g) + 3h_{2}(g) \leftrightarrow 2nh_{3}(g)}$

  • $k_{eq} = \frac{\ce{n_{2}}\ce{h_{2}}^{3}}{\ce{nh_{3}}^{2}}$
  • $k_{eq} = \frac{\ce{nh_{3}}^{2}}{\ce{n_{2}}\ce{h_{2}}^{3}}$
  • $k_{eq} = \frac{\ce{n_{2}}\ce{h_{2}}}{\ce{nh_{3}}}$
  • $k_{eq} = \frac{\ce{nh_{3}}}{\ce{n_{2}}\ce{h_{2}}}$
  • $k_{eq} = \ce{n_{2}}\ce{h_{2}}^{3}$

Explanation:

Question 4
Brief Explanations
  • For the equilibrium constant \( K_{eq} \):
  • Option A: \( K_{eq} \) is defined as the ratio of the concentrations (or partial pressures for gases) of products to reactants, each raised to their stoichiometric coefficients. So it represents the relative concentration of products compared to reactants. This is correct.
  • Option B: \( K_{eq} \) is products over reactants, not reactants over products. So this is incorrect.
  • Option C: \( K_{eq} \) is a dimensionless quantity (or has no units in the case of activities; for concentration - based \( K_{c} \), units can cancel out depending on the reaction, but it is not measured in moles). So this is incorrect.
  • Options D and E: Since B and C are incorrect, these are also incorrect.
Brief Explanations

The formula for the equilibrium constant \( K_{eq} \) (or \( K_{c} \) for concentration - based) for a reaction \( aA + bB
ightleftharpoons cC + dD \) is \( K_{eq}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \), where the brackets represent the molar concentrations of the species at equilibrium.

For the reaction \( \ce{N_{2}(g) + 3H_{2}(g)
ightleftharpoons 2NH_{3}(g)} \), the reactants are \( \ce{N_{2}} \) (with stoichiometric coefficient 1) and \( \ce{H_{2}} \) (with stoichiometric coefficient 3), and the product is \( \ce{NH_{3}} \) (with stoichiometric coefficient 2).

Using the formula for \( K_{eq} \), we have \( K_{eq}=\frac{[\ce{NH_{3}}]^{2}}{[\ce{N_{2}}][\ce{H_{2}}]^{3}} \).

Answer:

A. It represents the relative concentration of products compared with reactants.

Question 5