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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 15 of 73
Marks: +1, -0
At 700 K, equilibrium constant for the reaction : H2(g)+I2(g)H_{2(g)} + I_{2(g)}2Hl(g)2Hl_(g) is 54.8. If 0.5 molL1mol L^{–1} of HI(g)HI_(g) is present at equilibrium at 700 K, what are theconcentration of H2(g)H_{2(g)} and I2(g)I_{2(g)} assuming that we initially started with HI(g)HI_(g) and allowed it to reach equilibrium at 700 K?
Solution:  
2Hl(g)2Hl_(g)H2(g)+I2(g)H_{2(g)} + I_{2(g)}
Initial concentrationa00At equilibrium0.5xx\begin{array}{cccc} \text{Initial concentration} & a & 0 & 0 \\ \text{At equilibrium} & 0.5 & x & x \end{array}
KcK_c = [H2][I2]/[HI]2{[H_2][I_2]}/[HI]^2 or , 1/54.81/{54.8} = x×x/(0.5)2{x×x}/(0.5)^2 or x2x^2 = 0.25/54.8{0.25}/{54.8} ⇒ x = 0.068
[H2][H_2] = [I2][I_2] = 0.068 mollitre1mol litre^{-1}
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