# please if u can solve it and bid u will be reported or redrawed

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An equilibrium mixture, at 672°C in a 1155-mL container, involving the chemical system

N2(g) + 3H2(g)  2NH3(g)

is found to contain 9.14 g of N2, 0.707 g of H2, and 0.300 g of NH3. Calculate the equilibrium constant (Keq expressed in terms of the molar concentrations) at the given temperature.

A 5.00-L vessel contained 4.61×10-2 mol of gaseous PCl3, 5.22×10-2 mol of gaseous PCl5, and 6.83×10-2 mol of Cl2 gas at equilibrium at 222°C.
Calculate the value of Keq (expressed in terms of the molar concentrations) for the reaction

PCl3(g) + Cl2(g)  PCl5(g)

May 8th, 2015

DATA:

Mass of N2= 9.14g = (0.3264 mol*1000)/1155 =0.2826M

Mass of H2=0.707g =(0.3535 mol*1000)/1155 =0.3060M

Mass of NH3= o.300g = (0.01764 mol*1000)/1155=0.01527M

Keq =?

SOLUTION:

Keq = [NH3]2/[N2] [H2]3

By putting the molar concentrations of reactants and product in the above equation;

Keq = [.01527]2/[.2826] [0.3060]3

= 0.0287

DATA:
mol of PCl3 = (4.61*10-2 mol*1000)/5000=9.22*10-3 M

Mole of PCl5 = (6.83*10-2 mol *1000)/5000=1.366*10-2M

Mole of Cl2 = (5.22*10-2*1000)/5000= 1.044*10-2M

Keq =?

SOLUTION:

Keq = [PCl5]/[PCl3] [Cl2]

= 1.366*10-2/(9.22*10-3) (1.044*10-2)

Keq = 14.1912

May 8th, 2015

hello i entered both and they are incorrect

May 8th, 2015

DATA:
mol of PCl3 = (4.61*10-2 mol*1000)/5000=9.22*10-3 M

Mole of PCl5 = (6.83*10-2 mol *1000)/5000=1.366*10-2M

Mole of Cl2 = (5.22*10-2*1000)/5000= 1.044*10-2M

Keq =?

SOLUTION:

Keq = [PCl5]/[PCl3] [Cl2]

= 1.366*10-2/(9.22*10-3) (1.044*10-2)

Keq = 141.912

May 8th, 2015

it is wrong again

May 8th, 2015

i only found out mistake in this question

the question having NH3 is correct with the values you have provided me

May 8th, 2015

May 9th, 2015

i have given you the answer with solutions.

May 9th, 2015

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May 8th, 2015
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May 8th, 2015
Nov 24th, 2017
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