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Chemistry Question on Equilibrium

For the reaction 2HI(g)⇌H2(g)+I2(g)2 HI (g) \rightleftharpoons H _{2}(g)+ I _{2}(g) the degree of dissociation (α)(\alpha) of HI(g) is related to equilibrium constant KpK_p by the expression:

A

1+2Kp2\frac{1+2\sqrt{K_{p}}}{2}

B

1+2Kp2\sqrt{\frac{1+2K_{p}}{2}}

C

2Kp1+2Kp\sqrt{\frac{2K_{p}}{1+2K_{p}}}

D

2Kp1+2Kp\frac{2\sqrt{K_{p}}}{1+2\sqrt{K_{p}}}

Answer

2Kp1+2Kp\frac{2\sqrt{K_{p}}}{1+2\sqrt{K_{p}}}

Explanation

Solution

For the reaction :


∵\because Partial pressure == Mole fraction ×pT\times p_{T}

and Kp=pH2×pN2p(HH)2K_{p}=\frac{p_{ H _{2}} \times p_{ N _{2}}}{p_{( HH )}^{2}}
⇒pH2=(α2)pT\Rightarrow p_{ H _{2}}=\left(\frac{\alpha}{2}\right) p_{T}
pN2=(α2)pTp_{ N _{2}}=\left(\frac{\alpha}{2}\right) p_{T}
⇒ pHI=(1−α)pT\Rightarrow\, p_{ HI }=(1-\alpha) p_{T}

Hence, Kp=[(α2)⋅pT]2(1−α)2⋅pT2 K_{p}=\frac{\left[\left(\frac{\alpha}{2}\right) \cdot p_{T}\right]^{2}}{(1-\alpha)^{2} \cdot p_{T}^{2}}
⇒α1−α=2Kp\Rightarrow \frac{\alpha}{1-\alpha}=2 \sqrt{K_{p}}

On doing cross multiplication

α=2Kp(1−α)\alpha=2 \sqrt{K_{p}}(1-\alpha)

or 2Kp−2Kp⋅α=α2 \sqrt{K_{p}}-2 \sqrt{K_{p}} \cdot \alpha =\alpha
2Kp=α+2Kp2 \sqrt{K_{p}}=\alpha+2 \sqrt{K_{p}}
2Kp=α(1+2Kp)2 \sqrt{K_{p}}=\alpha\left(1+2 \sqrt{K_{p}}\right)
∴ α=2Kp1+2Kp\therefore\, \alpha =\frac{2 \sqrt{K_{p}}}{1+2 \sqrt{K_{p}}}