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  • The reaction, <img alt="2A(g) + B(g) \rightarrow 3C(g) + D(g)" src="https://entrancecorner.oncodecogs.c

The reaction,

2A(g) + B(g) \rightarrow 3C(g) + D(g)

is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression:

Option: 1

\frac{(0.75)^30.25}{(1.00)^21.00}


Option: 2

\frac{(0.75)^30.25}{(0.50)^2(0.75)}


Option: 3

\frac{(0.75)^30.25}{(0.50)^2(0.25)}


Option: 4

\frac{(0.75)^30.25}{(0.75)^2(0.25)}


Answers (1)

2A   +   B   =   3C   +   D

1            1           0         0          at t=0

1-2x        1-x     3x         x           at equilibrium

given x = 0.25

0.5           0.75   0.75     0.25

\begin{aligned} &\text { Equilibrium constant, } K=\frac{[C]^{3}[D]}{[A]^{2}[B]}\\ &\therefore \quad K=\frac{(0.75)^{3}(0.25)}{(0.5)^{2}(0.75)} \end{aligned}

The Correct answer is option 2. 

Posted by

Ramraj Saini

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