Question

2 questions

An empty steel container is filled with 0.830 atm of A and 0.830 atm of B. The system is allowed to reach equilibrium according to the reaction below. If Kp = 340 for this reaction, what is the equilibrium partial pressure of C? A (g) + B (g) = C (g) atm 1 2 3 Х c 7 8 9 +/- x 100 Tap here or pull up for additional resources

Answer 1

Question 1:

Using the ICE table

	A	B	C
Initial	0.830	0.830	0
Change	-x	-x	+x
Final	0.830-x	0.830-x	x

$K_p = \frac{[Products]}{[Reactants]} = \frac{[C]}{[A][B]} = \frac{x}{(0.830-x)^2} = 340$

Solving the quadratic equation we get

Hence the partial pressure of gas C at equilibrium will be 0.782 atm

Question 2:

Using the ICE table

	[HIO3]	[H+]	[IO3-]
Initial	0.570	0	0
Change	-x	+x	+x
Final	0.570-x	x	x

$K_c = \frac{[Products]}{[Reactants]} = \frac{[H^+][IO_3^-]}{[HIO_3]} = \frac{x^2}{0.570-x} = 0.170$

Solving the quadratic equation we get

r=0.238.1

[H+] = 0.238 M