Chemistry Equilibrium MDCAT 2017
PMDC Verified Question 85 of 102
For which of the following equilibrium reaction, \( K_c \) has no units?
A
\( \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \)
B
\( \text{CO}_{(g)} + \text{H}_2\text{O}_{(g)} \rightleftharpoons \text{CO}_{2(g)} + \text{H}_{2(g)} \)
C
\( 2\text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{SO}_{3(g)} \)
D
\( 2\text{NO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{NO}_{(g)} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( \text{CO}_{(g)} + \text{H}_2\text{O}_{(g)} \rightleftharpoons \text{CO}_{2(g)} + \text{H}_{2(g)} \)
Concept:

The unit of \( K_c \) is given by \( (\text{mol dm}^{-3})^{\Delta n} \). It will be unitless only if \( \Delta n = 0 \).

Formula:

$$ \Delta n = \Sigma \text{Moles of gaseous products} - \Sigma \text{Moles of gaseous reactants} $$

Solution:

  • For option B: \( \text{CO}_{(g)} + \text{H}_2\text{O}_{(g)} \rightleftharpoons \text{CO}_{2(g)} + \text{H}_{2(g)} \)


  • Reactant moles = \( 1 + 1 = 2 \). Product moles = \( 1 + 1 = 2 \).


  • \( \Delta n = 2 - 2 = 0 \). Therefore, units cancel out.


Why other options are incorrect:

Option A has \( \Delta n = 2 - 4 = -2 \). Option C has \( \Delta n = 2 - 3 = -1 \). Option D is unbalanced as written but represents \( \Delta n \neq 0 \).

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