Chemistry Equilibrium MDCAT 2014
PMDC Verified Question 95 of 102
The value of equilibrium constant \( K_c \) for the reaction \( 2\text{HF}_{(g)} \rightleftharpoons \text{H}_{2(g)} + \text{F}_{2(g)} \) is \( 10^{-13} \) at \( 2000^{\circ}\text{C} \). Calculate the value of \( K_p \) for this reaction
A
\( 2 \times 10^{-13} \)
B
\( 186 \times 10^{-13} \)
C
\( 10^{-13} \)
D
\( 3.48 \times 10^{-9} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( 10^{-13} \)
Concept:

The relationship between \( K_p \) and \( K_c \) is governed by the change in gaseous moles (\( \Delta n \)) during the reaction.

Formula:

$$ K_p = K_c(RT)^{\Delta n} $$

Solution:

  • Calculate \( \Delta n \) for \( 2\text{HF}_{(g)} \rightleftharpoons \text{H}_{2(g)} + \text{F}_{2(g)} \).


  • Moles of gaseous products = \( 1 + 1 = 2 \).


  • Moles of gaseous reactants = 2.


  • \( \Delta n = 2 - 2 = 0 \).


  • Therefore, \( K_p = K_c(RT)^0 = K_c(1) = K_c \).


  • Since \( K_c = 10^{-13} \), \( K_p \) must also be \( 10^{-13} \).


Why other options are incorrect:

Other values would only be correct if \( \Delta n \neq 0 \) and the term \( (RT)^{\Delta n} \) altered the value of \( K_c \).

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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