Question

part b.

Determine the equilibrium constant, *K*goal, for the
reaction

4PCl5(g)⇌P4(s)+10Cl2(g), *K*goal=?

by making use of the following information:

P4(s)+6Cl2(g)⇌4PCl3(g), *K*1=2.00×1019

PCl5(g)⇌PCl3(g)+Cl2(g), *K*2=1.13×10−2

Express your answer numerically.

Answer #1

**Solution :-**

In the final equation we have P4 on the product side therefore we need to reverse the equation 1

And multiply equation 2 by 4 to get the 4 PCl5

After reversing the equation the Kc for the new equation is the inverse of the K1

And multiplying second equation by 4 means k for the new eequation is (K2)^4

Then add both equations by cancelling the similar species on the opposite sides and the K values are multipled.

Therefore new equations are as follows

4PCl3(g) ⇌
P4(s)+6Cl2(g)
1/*K*1=1/2.00×1019

4PCl5(g)⇌4PCl3(g)+4Cl2(g),
(*K*2)^4=(1.13×10−2)^4

___________________________________________________

4PCl5(g) ⇌P4(s) + 10
Cl2(g)
K_{goal} = (1/K1)*(K2)^4

Therefore K_{goal} = (1/2.00×1019)*(1.13*10^-2)^4 =
8.15*10^-28

**Therefore equilibrium constant K _{goal} =
8.15*10^-28**

Determine the equilibrium constant, Kgoal, for the
reaction 4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=?
by making use of
the following information:
P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019
PCl5(g)⇌PCl3(g)+Cl2(g), K2=1.13×10−2

What is the enthalpy change for the first reaction?
P4(s) + 6Cl2(g) → 4PCl3(l) ΔH
=
P4(s) + 10Cl2(g) → 4PCl5(s) ΔH
= -1,779.8
PCl3(l) + Cl2 → PCl5(s) ΔH =
-123.3
question 2
What is the enthalpy change for the first reaction?
Fe2O3(s) → 2Fe(s) + 3/2O2(g) ΔH
=
4Fe(s) + 3O2(g) → 2Fe2O3 (s) ΔH
= -1,645 kJ
Help me understand, please

P4(s) + 6Cl2(g) ==> 4PCl3(l)
If 112g of P4 and 303g of Cl2 react what mass of PCl3 is
expected to form? What is the limiting reactant?

Elemental phosphorus reacts with chlorine gas according to the
equation: P4(s)+6Cl2(g)→4PCl3(l) A reaction mixture initially
contains 45.21 g P4 and 130.4 g Cl2.
Part A Once the reaction has occurred as completely as possible,
what mass (in g) of the excess reactant remains?
Express your answer to three significant figures.
m =

what is the delta H°reaction for the reaction shown
below when 54.2g of Cl2 react? P4(s) + 6Cl2(g) ---> 4PCl3(g)
delta H = -1226kj

Determine the value of the equilibrium constant, Kgoal, for the
reaction N2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=? by making use of the
following information:
1. N2(g)+O2(g)⇌2NO(g), K1 = 4.10×10−31
2. N2(g)+2H2(g)⇌N2H4(g), K2 = 7.40×10−26
3. 2H2O(g)⇌2H2(g)+O2(g), K3 = 1.06×10−10

The equilibrium constant, Kc, for the following reaction is 83.3
at 500 K. PCl3(g) + Cl2(g) PCl5(g) Calculate the equilibrium
concentrations of reactant and products when 0.269 moles of PCl3
and 0.269 moles of Cl2 are introduced into a 1.00 L vessel at 500
K. [PCl3] = M [Cl2] = M [PCl5] = M

For the exothermic reaction
PCl3(g)+Cl2(g)?PCl5(g)
Kp = 0.200 at a certain temperature.
A flask is charged with 0.500 atm PCl3 , 0.500
atm Cl2, and 0.300 atm PCl5 at this temperature.
What are the equilibrium partial pressures of PCl3
, Cl2, and PCl5, respectively?
Express your answers numerically in atmospheres with three
digits after the decimal point, separated by commas.

The equilibrium constant, Kc, for the following reaction is
1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) Calculate the
equilibrium concentrations of reactant and products when 0.390
moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K.

The equilibrium constant KC KC for the reaction
PCL3+ CL2 is in equilibrium yielding PCL 5
equals 2.7 at 330°C.
a
A sample of 37.0 g of PCL5 is placed in a 2.1 L reaction vessel
and heated to 330°C. What are the equilibrium concentrations of all
of the species?
PCL5= M
PCL3
CO2= M

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