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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 17 of 73
Marks: +1, -0
KpK_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6\mathrm{C_2H_6} when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
C2H6(g)\mathrm{C_2H_{6(g)}} ⇌ C2H4(g)+H2(g)\mathrm{C_2H_{4(g)} + H_{2(g)}}
Solution:  
C2H6(g)\mathrm{C_2H_{6(g)}} ⇌ C2H4(g)+H2(g)\mathrm{C_2H_{4(g)} + H_{2(g)}}
Initial pressure4.0 atm00At. eqm.4−ppp\begin{array}{lccc} \text{Initial pressure} & 4.0\ \text{atm} & 0 & 0 \\ \text{At. eqm.} & 4 - p & p & p \end{array}
Applying law of chemical equilibrium,
KpK_p = pC2H4×pH2pC2H6\frac{p_{C_2H_4} \times p_{H_2}}{p_{C_2H_6}} ⇒ 0.041 = p24−p\frac{p^2}{4-p} or , p2p^2 = 0.16 – 0.04p or p2p^2 + 0.04p – 0.16 = 0
∴ p = −0.04±0.0016−4(−0.16)2\frac{-0.04 \pm \sqrt{0.0016 - 4(-0.16)}}{2} = −0.04±0.802\frac{-0.04 \pm 0.80}{2}
Taking positive value, p = 0.762\frac{0.76}{2} = 0.38
∴ [C2H6]eq[\mathrm{C_2H_6}]_{\text{eq}} = 4 - 0.38 atm = 3.62 atm
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