Question & Answer: At temperatures near 800 degree C, steam passed over Hot Coke reacts to form Carbon…..

At temperatures near 800°C, steam passed over Hot Coke reacts to form Carbon Monoxide and Hydrogen: 4. .. C(s) + H2O(g) CO(g) + H2(g) r The mixture of gases that results is an important industrial fuel called Water Gas. a) At 800°C Kp 14.1. What are the equilibrium partial pressures of each gas in the equilibrium mixture at this temperature if we start with 0.10 mole Coke and 0.10 mole Steam in a Lvessel? b) What is the minimum amount of Coke required to achieve equilibrium under these conditions? c What is the total pressure in the vessel at equilibrium? d) At 25°C he value of Kp for this reaction is 1.7 x 1021 Is the reaction Exothermic or Endothermic? To produce the maximum amount of product, should the pressure of the system be increased or decreased? e)

At temperatures near 800 degree C, steam passed over Hot Coke reacts to form Carbon Monoxide and Hydrogen: C_(s) + H_2O_(g) CO_(g) + H_2_(g) The mixture of gases that results is an important industrial fuel called Water Gas. a) At 800 degree C K_p = 14.1. What are the equilibrium partial pressures of each gas in the equilibrium mixture at this temperature if we start with 0.10 mole Coke and 0.10 mole Steam in a 1 L vessel? b) What is the minimum amount of Coke required to achieve equilibrium under these conditions? c) What is the total pressure in the vessel at equilibrium? d) At 25 degree C he value of K_p for this reaction is 1.7 times 10^-21 Is the reaction Exothermic or Endothermic? e) To produce the maximum amount of product, should the pressure of the system be increased or decreased?

Expert Answer

Answer

a) Initial number of moles of gaseous species (steam) $=0.10$

Initial total pressure $P = dfrac {nRT}{V} = dfrac {0.10 : mol times 0.08206 : L atm/mol K times (800+273) : K }{1 : L} = 8.81 ; atm$

	$H_2O$	$CO$	$H_2$
Initial pressure (atm)	8.81	0	0
Change in pressure (atm)	-x	x	x
Equilibrium pressure (atm)	8.81-x	x	x

The equilibrium constant

$K_p = dfrac {P_{CO}P_{H2}}{P_{H2O}}$

$14.1 = dfrac { x times x}{8.81-x}$

$8.81-x = dfrac { x times x}{14.1}$

$8.81-x = 0.0709x^2$

$0.0709x^2-x-8.81=0$

This is quadratic equation with solutions

$x= dfrac {-b pm sqrt {b^2-4ac}}{2a}$

$x= dfrac {-1 pm sqrt {1^2-4 times 0.0709 times (-8.81)}}{2 times 0.0709}$

$x= dfrac {-1 pm 1.8704}{0.1418}$

$x= -2.87$

$x=6.139$

The negative value is discarded as pressure cannot be negative.

The equilibrium partial pressures of CO and H2 are 6.139 atm. The equilibrium partial pressure of
$H_2O = 8.81-x=8.81-6.139=2.67 : atm$ .

b) Out of 8.81 atm, 6.139 atm of steam are consumed to reach equilibrium. This corresponds to $0.10 : mol times dfrac {6.139 : atm}{8.81 : atm} = 0.0697 : mol$ .

Hence, the minimum amount of coke required is 0.0697 moles.

c) Total pressure in the vessel at equilibrium $=8.81-x +x +x = 8.81+x = 8.81+6.139=14.9 : atm$

d) At 800 deg C, $K_p = 14.1$

At 25 deg C, $K_p = 1.7 times 10^{-21}$

When temperature is increased from 25 deg C to 800 deg C, the value of Kp also increases from $1.7 times 10^{-21}$ to 14.1. This indicates endothermic reaction which involves absorption of heat.

e) To produce the maximum amount of the product, the pressure of the system should be decreased. The reaction proceeds with increase in the number of moles of gaseous species.

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