Chemical EquilibriumHard

Question

For the reversible reaction, N2(g) + 3H2(g) ­$\rightleftharpoons$2NH3(g) at 500°C, the value of Kp is 1.44 × 10–5 when partial pressure is measured in atmospheres. The corresponding value of Kc, with concentration in mole litre–1 is

Options

A.$\frac{1.44 \times 10^{- 5}}{(0.082 \times 500)^{- 2}}$
B.$\frac{1.44 \times 10^{- 5}}{(8.314 \times 773)^{- 2}}$
C.$\frac{1.44 \times 10^{- 5}}{(8.314 \times 500)^{- 2}}$
D.$\frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}}$

Solution

Δng = 2 – (1 + 3) = –2

$K_{p} = K_{c}.(RT)_{g}^{\Delta n} = K_{c}.(RT)^{- 2} $$$\therefore K_{c} = \frac{K_{p}}{(RT)^{- 2}} = \frac{1.44 \times 10^{- 5}}{(0.0821 \times 773)^{- 2}}$$

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