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The ratio $\frac{K_P}{K_C}$ for the reaction: $CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$ is:
The ratio $\frac{K_P}{K_C}$ for the reaction: $CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$ is:
- $\frac{1}{\sqrt{RT}}$ (correct)
- 1
- RT
- $(RT)^{1/2}$
Flashcards
K
P
K P
The ratio of partial pressures of products to reactants at equilibrium.
K
C
K C
The ratio of molar concentrations of products to reactants at equilibrium.
K
P and K
C Relation
K P and K C Relation
K P = K C(RT)^\Delta n, where \Delta n is the change in moles of gas.
Calculating Δn
Calculating Δn
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Change in moles (\Delta n)
Change in moles (\Delta n)
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\Delta n for given reaction
\Delta n for given reaction
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K
P / K
C ratio
K P / K C ratio
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Study Notes
- The question pertains to JEE Main 2025 exam for Chemistry.
- It gives 4 marks for a correct answer and -1 for a wrong answer.
- The question asks about the ratio of Kp/Kc.
- The reaction is: CO(g) + 1/2 O2(g) <=> CO2(g)
Finding Kp/Kc
- Kp/Kc = (RT)^Δn
- Δn is the change in the number of moles of gas.
- Δn = (moles of gaseous products) - (moles of gaseous reactants)
- Δn = 1 - (1 + 1/2) = 1 - 3/2 = -1/2
- Kp/Kc = (RT)^(-1/2) = 1 / √(RT)
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