JEE Advance - Chemistry (2016 - Paper 2 Offline - No. 5)
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Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equations:
X2 (g) $$\leftrightharpoons$$ 2X (g)
The standard reaction Gibbs energy, $$\Delta _rG^o$$, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by $$\beta$$. Thus, $$\beta _{equilibrium}$$ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given R = 0.083 L bar K-1 mol-1)
Question
The INCORRECT statement among the following for this reaction, is
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equations:
X2 (g) $$\leftrightharpoons$$ 2X (g)
The standard reaction Gibbs energy, $$\Delta _rG^o$$, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by $$\beta$$. Thus, $$\beta _{equilibrium}$$ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given R = 0.083 L bar K-1 mol-1)
Question
The INCORRECT statement among the following for this reaction, is
Decrease in the total pressure will result in formation of more moles of gaseous X
At the start of the reaction, dissociation of gaseous X2 takes place spontaneously
$${{\beta _{equilibrium}}}$$ = 0.7
Kc < 1
Explanation
On decreasing total pressure, the reaction will move in the forward direction where number of gaseous molecules is less.
Hence, option (A) is correct.
At the start of reaction Q = 0. From $$\Delta$$G = $$\Delta$$Go + RT lnQ, we have that for Q = 0 at the start of reaction $$\Delta$$rG is negative, this causes dissociation of X2 to take place spontaneously.
Hence, option (B) is correct.
If bequilibrium = 0.7, then value of equilibrium constant is
$${K_p} = {{8\beta _{eq}^2} \over {4 - \beta _{eq}^2}} = {{8{{(0.7)}^2}} \over {7 - {{(0.7)}^2}}}$$
The value of Kp is greater than 1, which is not possible as given that $$\Delta$$Go > 0 for the reaction.
Hence, option (C) is incorrect.
As $$\Delta$$Go > 0 and $$\Delta$$Go = $$-$$RT ln(Kp)
$$\Delta$$Go > 1, so Kp should be less than 1.
We know that
$${K_p} = {K_c}{(RT)^{\Delta {n_g}}}$$
$${K_c} = {{{K_p}} \over {{{(RT)}^{\Delta {n_g}}}}}$$
Therefore, Kc < Kp . As Kp is less than 1, so Kc is also less than 1.
Hence, option (D) is correct.
Hence, option (A) is correct.
At the start of reaction Q = 0. From $$\Delta$$G = $$\Delta$$Go + RT lnQ, we have that for Q = 0 at the start of reaction $$\Delta$$rG is negative, this causes dissociation of X2 to take place spontaneously.
Hence, option (B) is correct.
If bequilibrium = 0.7, then value of equilibrium constant is
$${K_p} = {{8\beta _{eq}^2} \over {4 - \beta _{eq}^2}} = {{8{{(0.7)}^2}} \over {7 - {{(0.7)}^2}}}$$
The value of Kp is greater than 1, which is not possible as given that $$\Delta$$Go > 0 for the reaction.
Hence, option (C) is incorrect.
As $$\Delta$$Go > 0 and $$\Delta$$Go = $$-$$RT ln(Kp)
$$\Delta$$Go > 1, so Kp should be less than 1.
We know that
$${K_p} = {K_c}{(RT)^{\Delta {n_g}}}$$
$${K_c} = {{{K_p}} \over {{{(RT)}^{\Delta {n_g}}}}}$$
Therefore, Kc < Kp . As Kp is less than 1, so Kc is also less than 1.
Hence, option (D) is correct.
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