Question

I need all answers please

2) Mustard gas, used in chemical warfare in World War I, has been found to be an effective agent in the chemotherapy of Hodgkin's disease. It can be produced according to the following reaction: SCl (g)+ 2C H, (g)S(CH2CH2CI)2 (g) An evacuated 5.0 L flask at 20.0°C is filled with 0.258 mol SC12 and 0.592 mol C2H4. After equilibrium is established, 0.0349 mol mustard gas is present. (a) What is the partial pressure of each gas at equilibrium? (b) What is K at 20.0°C?

Answer 1

Given, the balanced chemical reaction,

SCl₂(g) + 2C₂H₄(g) $\rightleftharpoons$ S(CH₂CH₂Cl)₂(g)

Also given,

The volume of the flask(V) = 5.0 L

Temperature(T) = 20.0 ^oC + 273.15 = 293.15 K

Number of moles of SCl₂(g) = 0.258 mol

Number of moles of C₂H₄(g) = 0.592 mol

Number of moles of S(CH₂CH₂Cl)₂(g) at equilibrium = 0.0349 mol

Firstly,calculating the partial pressures of each given quantity,

We know, the ideal gas law formula,

PV = nRT

Rearranging the formula,

P = nRT /V

P_SCl2 = (0.258 mol x 0.08206 L.atm/mol.K x 293.15 K) / 5.0 L

P_SCl2 = 1.24 atm

Similarly,

P_C2H4 = (0.592 mol x 0.08206 L.atm/mol.K x 293.15 K) / 5.0 L

P_C2H4 = 2.848 atm

Also,

P_S(CH2CH2Cl)2(equiiibrium) = (0.0349 mol x 0.08206 L.atm/mol.K x 293.15 K) / 5.0 L

P_S(CH2CH2Cl)2(equiiibrium) = 0.168 atm

Drawing an ICE chart,

	SCl2(g)	2C2H4(g)	S(CH2CH2Cl)2(g)
I(atm)	1.241	2.848	0
C(atm)	-x	-2x	+x
E(atm)	1.241-x	2.848-2x	x

Now,

a) Partial pressure of each gas at equilibrium,

P_SCl2(eq) = 1.241- x = 1.073 atm Or 1.1 atm [2 S.F]

P_C2H4(eq) = 2.848-2x = 2.512 atm Or 2.5 atm [2 S.F]

P_{S(CH2CH2Cl)2(g)}(eq) = x = 0.168 atm Or 0.17 atm [2 S.F]

b) Now, the equilbrium constant is,

K_p = P_{S(CH2CH2Cl)2(g)} / [P_SCl2(g) x P_C2H4(g)² ]

K_p = 0.168 / [1.073 x 2.512² ]

K_p = 0.0248 Or 0.025 [2 S.F]