Chemistry MCQ thread...

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When solid ammonium chloride dissociates at a certain temperature in a 0.500dm3 container, ammonia and hydrogen chloride are formed. NH4Cl(s) NH3(g) + HCl(g) The initial amount of ammonium chloride was 1.00mol, and when the system had reached equilibrium there was 0.300mol of ammonium chloride. What is the numerical value of Kc for this reaction under these conditions?
A 0.490 B 1.63 C 1.96 D 3.27
can u show me how to sove it??
Note that ammonium chloride does not appear in the Kc expression as it is a SOLID.
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The equation constant for the following reaction is 1.39 ×10^-2
2HI ⇌ H^2 + I^2
What will be the equilibrium constant for the following reaction?
H^2+ I^2 ⇌ 2HI
How do i solve this??
 
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The equation constant for the following reaction is 1.39 ×10^-2
2HI ⇌ H^2 + I^2
What will be the equilibrium constant for the following reaction?
H^2+ I^2 ⇌ 2HI
How do i solve this??
Won't it be the reciprocal of 1.39 x 10^-2 i.e. 71.9?
 
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Won't it be the reciprocal of 1.39 x 10^-2 i.e. 71.9?
I don't think it will be reciprocal ....when writing initial moles and the final moles of the first equation
x2
(1-2x)2
=1.39*10-2

for the next reaction
(2x)2
(1-x)(1-x)
=?
and i am getting Kc=4.51*10-2
 
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884
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I don't think it will be reciprocal ....when writing initial moles and the final moles of the first equation
x2
(1-2x)2
=1.39*10-2

for the next reaction
(2x)2
(1-x)(1-x)
=?
and i am getting Kc=4.51*10-2
When u take the moles for any product to be x in the first reaction, then y don't u take it as x in the next reaction, where it becomes the moles of the reactant?
 
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When u take the moles for any product to be x in the first reaction, then y don't u take it as x in the next reaction, where it becomes the moles of the reactant?
Taking it as X in the reactants of the reaction will show that the number of moles at equilibrium and before equilibrium are same. In order to obtain moles at equilibrium we subtract the initial moles, which is considered to be '1' from the unknown decrease stated as 'X' after equilibrium is reached (1-x). However product will be taken as '2X' after equilibrium is reached.
 
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Please hhelp with this question for may june 2006

Use of the Data Booklet is relevant to this question.

What mass of solid residue can be obtained from the thermal decomposition of 4.10

g of
anhydrous calcium nitrate?

A 0.70

g B 1.00g

C 1.40g

D 2.25g
 
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Please hhelp with this question for may june 2006

Use of the Data Booklet is relevant to this question.

What mass of solid residue can be obtained from the thermal decomposition of 4.10

g of
anhydrous calcium nitrate?

A 0.70

g B 1.00g

C 1.40g

D 2.25g
Moles of calcium nitrate= 4.10/164.1 = 0.02498
Moles of calcium oxide= 0.02498
Mass of calcium oxide= 0.02498 x 56.1=1.40g
 
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