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Units and Dimensions

KP KC

Question

$K_{P}$ & $K_{C}$

A

k_{C} = 197.85 m o l^{2} l i t^{2}, k_{P} = 0.05 a t m^{2}

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B

k_{C} = 297.95 m o l^{2} l i t^{2}, k_{P} = 0.1 a t m^{2}

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C

k_{C} = 300 m o l^{2} l i t^{2}, k_{P} = 0.15 a t m^{2}

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D

None of these

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Solution

The correct option is A $k_{C} = 197.85 m o l^{2} l i t^{2}, k_{P} = 0.05 a t m^{2}$

${C O}_{(g)} + 2 H_{2 (g)} ⇌ {C H}_{3} {O H}_{(g)}$

Initially $0.15$
$0.15 - 0.08 0.08$
___________________
0.07/volume x/vol 0.08/vol

$n = \frac{P V}{R T} = \frac{8.5 \times 2.5}{0.0821 \times 750} = 0.34$

$0.07 + x + 0.08 = 0.34$

$\Rightarrow x = 0.19$

$K_{c} = \frac{[C H_{3} O H]}{[CO (H_{2})]^{2}} = \frac{0.08 / 2.25}{\frac{0.07}{2.5} \times {(\frac{0.19}{2.5})}^{2}}$

$K_{c} = 197.86 {m o l}^{- 2} L^{2}$

$K_{p} = K_{c} \times (R T)^{Δ n}$

$= 197.86 \times (0.082 \times 750)^{- 2}$

$= 0.052 {a t m}^{2}$

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