The reaction, 2 A(g)+B(g) leftharpoons 3 C(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

Question

The reaction, 2 A(g)+B(g) leftharpoons 3 C(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 (A) [(0.75)3(0.25)] ÷[(1.00)2(1.00)] (B) [(0.75)3(0.25)] ÷[(0.50)2(0.75)] (C) [(0.75)3(0.25)] ÷[(0.50)2(0.25)] (D) [(0.75)3(0.25)] ÷[(0.75)2(0.25)]

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/ce89bf7f4da7ff7af999ab4b072d6adc-.png" /> K=((0.75)3(0.25)/(0.50)2(0.75))

$[(0.75)^{3} (0.25)] \div [(1.00)^{2} (1.00)]$

$[(0.75)^{3} (0.25)] \div [(0.50)^{2} (0.75)]$

$[(0.75)^{3} (0.25)] \div [(0.50)^{2} (0.25)]$

$[(0.75)^{3} (0.25)] \div [(0.75)^{2} (0.25)]$

$K = \frac{( 0.75 ) ^{3} ( 0.25 )}{( 0.50 ) ^{2} ( 0.75 )}$

Q. The reaction, 2A(g)+B(g)⇌3C(g)+D(g) is begun with the concentrations of A and B both at an initial value of 1.00M. When equilibrium is reached, the concentration of D is measured and found to be 0.25M. The value for the equilibrium constant for this reaction is given by the expression

[(0.75)3(0.25)]÷[(1.00)2(1.00)]

[(0.75)3(0.25)]÷[(0.50)2(0.75)]

[(0.75)3(0.25)]÷[(0.50)2(0.25)]

[(0.75)3(0.25)]÷[(0.75)2(0.25)]

K=(0.50)2(0.75)(0.75)3(0.25)​

$[(0.75)^{3} (0.25)] \div [(1.00)^{2} (1.00)]$

$[(0.75)^{3} (0.25)] \div [(0.50)^{2} (0.75)]$

$[(0.75)^{3} (0.25)] \div [(0.50)^{2} (0.25)]$

$[(0.75)^{3} (0.25)] \div [(0.75)^{2} (0.25)]$

$K = \frac{( 0.75 ) ^{3} ( 0.25 )}{( 0.50 ) ^{2} ( 0.75 )}$