1)at 227 degrees celcius, carbon monoxide gas reacts with hydrogen gas to form o

Question

1)at 227 degrees celcius, carbon monoxide gas reacts with hydrogen gas to form one mole of methanol, CH3OH (g). At equilibrium, the partial pressures of the gases are Pco=0.295 atm, Ph2=1.25 atm, and Pch3oh=2.18atm
a)write a balanced equation for the reaction
b)calculate the equilibrium constant for the reaction
2)consider the reaction:
A+B <====> (reversible equation icon) C+3D
A solution was prepared by mixing 50.00 mL of 1.00×10^-3 M A, and 100mL of 2.00×10^-3 M B, 10.0 mL of 1.0 M C and 75 ml of 1.50×10^-3 M D. At equilibrium, the concentration of D was measured to be 6.0×10^-4 M
a)calculate the equilibrium concentrations of each species
b)calculate the equilibrium constant
3)consider the reaction:
2NO2 (g) + CL2 (g) <====> (reversible equation icon) 2NOCL (g)
The initial pressure of NO2 was 1.577 atm and the initial pressure of Cl2 was 0.427 atm. At equilibrium, the partial pressure of NPCL is 0.624 atm. calculate K.

Explanation / Answer

1]

CO + 2H2 -----> CH3OH

K = pCH3OH / pH2^2 * pCO = 2.18 / 1.25^2 *0.295 = 4.729

2]

A + B ------> C + 3D

Concentrations after mixing [Moles / Total Volume ]

[A] = 1*10^-3 *50 / 50+100+10+75 = 2.127*10^-4 M

Similarly

[B] = 8.51*10^-4 M

[C] = 0.04255 M

[D] = 4.78*10^-4 M

A + B ------> C + 3D

[A] [B] [C] [D]

[A]-x [B]-x [C]+x [D]+3x

[D]+3x = 6*10^-4

x = 6*10^-4 - 4.78*10^-4 = 1.22*10^-4 /3 = 4.067*10^-5

[A] = 1.72*10^-4 M

[B] = 8.1033*10^-4 M

[C] = 0.04259

K = [C] [D]^3 / [A] [B] = 6.6*10^-5

3]

2NO2 + Cl2 ----> 2NOCl

1.577 0.427 0

1.577-2x 0.427-x 2x

2x = 0.624

x = 0.312

pNO2 = 1.577-0.624 = 0.953 atm

pCl2 = 0.115 atm

K = pNOCl^2 / pNO2^2 *pCl2 = 3.728

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