Question

Suppose a 250. mL flask is filled with 1.7 mol of H2S, 2.0 mol of CS2 and 0.50 mol of H 2. This reaction becomes possible: CH_(8) +2H2S(g) = CS2(g) + 4H2 () Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of CH4. You can leave out the M symbol for molarity. CH, HS CS H 8 initial Х 5 ? change x equilibrium

Answer 1

Given :

Volume of flask = 250 ml = 0.250 L

initial moles of H2S = 1.7 mol

initial moles of CS2= 2.0 mol

initial moles of H2 = 0.5 mol

Now as

$molarity = \frac{no.\;of\;moles}{volume \;in \;L}$

Thus

molarity of H2S 1.7 mol 0.250 L 6.8 mol/L

$molarity\;of\;CS_{2} = \frac{2\:mol}{0.250\:L}=8\:mol/L$

$molarity\;of\;H_{2} = \frac{0.5\:mol}{0.250\:L}=2\:mol/L$

As given molarity of CH4 could be repreesentes as M

Thus ,

$CH_{4}(g)+2H_{2}S(g)\rightarrow CS_{2}(g)+4H_{2}(g)$

	$CH_{4}$	$H_{2}S$	$CS_{2}$	$H_{2}$
initial	M	6.8	8	2
change	x	-2x	+x	+4x
equilibrium	M - x	6.8 - 2x	8 + x	2 + 4x

2. Given:

$N_{2}(g)+2H_{2}O(g)\rightarrow 2NO(g)+2H_{2}(g)$

Equilibrium constant in terms of pressure is the ratio of product of pressure of products raised to power equal to their stoichiometric coefficient to the  product of pressure of reactants raised to the power equal to their stoichiometric coefficient

Thus equilibrium constant in terms of pressure is given by ,

$K_{p}=\frac{(P_{NO})^{2}(P_{H_{2}})^{2}}{(P_{N_{2}})(P_{H_{2}O})^{2}}$