Solid carbon can react with gaseous water to form carbon monoxide gas and hydrog

Question

Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×103.

Part A

If a 1.55-L reaction vessel initially contains 194 torr of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.

Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×103.

Part A

If a 1.55-L reaction vessel initially contains 194 torr of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.

Explanation / Answer

C(s) + H2O (g) <--> CO (g) + H2 (g)   Kp = 0.0016

Initially we had P = 194 torr = 194 /760 atm = 0.25526 atm

V = 1.55 L , T = 700 K , we find moles of H2O using PV = nRT equation where R is gas constant

0.25526 atm x 1.55 L = n x 0.08206 liter atm/molK x 700 K

n = 0.00689

we find Kc of reaction , Kp = K ( RT)^dn

where dn = sum of coeffients of gas products - sum of coeffients of gas reactanats = 2-1 = 1

0.0016 = Kc ( 0.08206 x 700) ^1

Kc = 2.785 x 10^-5

now at equilibrium H2O moles = 0.00689 - X , CO moles= H2 moles = X

[H2O] = ( 0.0689-X) /1.55 , [CO] = [H2] = X/1.55

Kc = [CO] [H2] /[H2O]

2.785 x 10^-5 = (X^/1.55)^2 / ( 0.0689-X) /1.55

2.785 x 10^-5 x 1.55 = X^2 / ( 0.0689-X)    ( we approximate 0.0689-X neary 0.0689 since Kc is small and we get X value small)

X = 0.001725 = H2 moles

H2 mass = moles x molar mass of H2 = 0.001725 mol x 2 g/mol = 0.00345g

CO mass = 0.001725 x 28 g/mol = 0.0483 g

H2O mass = (0.0689-0.001725) x 18 g/mol = 1.2 g

total gas mass = 0.00345 + 0.0483 + 1.2 = 1.26 g

% of H2 = ( 100 x H2 mass) / ( total gas mass) = ( 100 x 0.00345 g / 1.26 g) = 0.274 %

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