-36. Consider the following reaction having an equilibrium constant of 4.66 x 10-3 at a certain temperature: A(g) + B(g) = C(g) If 0.300 mol of A and 0.100 mol of B are mixed in a container and allowed to reach equilibrium, the concentrations of A= 0.300 – x and B=0.100 – x reaction that form the concentration of C = 2x are related to the equilibrium constant by the expression (2x)² (0.300 – x) (0.100 – x)' 4.66 x 10-3 = (a) Write the quadratic equation for x. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula. (c) Find the cquilibrium concentration AR and C

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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2-36. Consider the following reaction having
an equilibrium constant of 4.66 x 10-3
at a certain temperature:
A(g) + B(g) = C(g)
If 0.300 mol of A and 0.100 mol of B
are mixed in a container and allowed to
reach equilibrium, the concentrations
of A= 0.300 – x and B=0.100 – x
reaction that form the concentration of
C = 2x are related to the equilibrium
constant by the expression
(2x)²
4.66 x 10-3
(0.300 – x) (0.100 – x)'
(a) Write the quadratic equation for x.
(b) Solve the quadratic equation found
in part (a) by completing the square
and the quadratic formula.
(c) Find the equilibrium concentration
A, B, and C.
Transcribed Image Text:2-36. Consider the following reaction having an equilibrium constant of 4.66 x 10-3 at a certain temperature: A(g) + B(g) = C(g) If 0.300 mol of A and 0.100 mol of B are mixed in a container and allowed to reach equilibrium, the concentrations of A= 0.300 – x and B=0.100 – x reaction that form the concentration of C = 2x are related to the equilibrium constant by the expression (2x)² 4.66 x 10-3 (0.300 – x) (0.100 – x)' (a) Write the quadratic equation for x. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula. (c) Find the equilibrium concentration A, B, and C.
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