JEE Main · 2024 · Shift-IIeasyCEQ-003

The ratio KPKC for the reaction: CO(g) + 12O2(g) CO2(g) is:

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The ratio KPKC\frac{K_P}{K_C} for the reaction: CO(g)+12O2(g)CO2(g)\mathrm{CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}} is:

Options
  1. a

    1RT\frac{1}{\sqrt{RT}}

  2. b

    (RT)1/2(RT)^{1/2}

  3. c

    RTRT

  4. d

    11

Correct Answera

1RT\frac{1}{\sqrt{RT}}

Detailed Solution

Formula: KP=KC(RT)ΔngK_P = K_C (RT)^{\Delta n_g}, so KPKC=(RT)Δng\frac{K_P}{K_C} = (RT)^{\Delta n_g}

Step 1 — Calculate Δng\Delta n_g: Δng=moles of gaseous productsmoles of gaseous reactants=1(1+12)=11.5=12\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - \left(1 + \frac{1}{2}\right) = 1 - 1.5 = -\frac{1}{2}

Step 2 — Calculate ratio: KPKC=(RT)1/2=1RT\frac{K_P}{K_C} = (RT)^{-1/2} = \frac{1}{\sqrt{RT}}

Answer: Option (1) 1RT\frac{1}{\sqrt{RT}}

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The ratio KPKC for the reaction: CO(g) + 12O2(g) CO2(g) is: (JEE Main 2024) | Canvas Classes