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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 11 of 73
Marks: +1, -0
A sample of HI(g)\mathrm{HI}_{(g)} is placed in flask at pressure of 0.2 atm. At equilibrium the partial pressure of HI(g)\mathrm{HI}_{(g)} is 0.04 atm. What is KpK_p for the given equilibrium?
2HI(g)2\mathrm{HI}_{(g)} ⇌ H2(g)+I2(g)\mathrm{H}_{2(g)}+\mathrm{I}_{2(g)}
Solution:  
2HI(g)2\mathrm{HI}_{(g)} ⇌ H2(g)+I2(g)\mathrm{H}_{2(g)}+\mathrm{I}_{2(g)}
Initial pressure0.2 atm00Pressure at eqm.0.04 atm0.162=0.08 atm0.162=0.08 atm\begin{array}{cccc} \text{Initial pressure} & 0.2\,\text{atm} & 0 & 0 \\ \text{Pressure at eqm.} & 0.04\,\text{atm} & \frac{0.16}{2} = 0.08\,\text{atm} & \frac{0.16}{2} = 0.08\,\text{atm} \end{array}
According to law of chemical equilibrium,
KpK_p = pH2×pI2pHI2\frac{p_{\mathrm{H}_2} \times p_{\mathrm{I}_2}}{p^2_{\mathrm{HI}}} = 0.08×0.08(0.04)2\frac{0.08 \times 0.08}{(0.04)^2} = 0.00640.0016\frac{0.0064}{0.0016} = 4
∴ KpK_p = 4
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