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3 years ago

Initially, [NH3(g)] = [O2(g)] = 3.60 M; at equilibrium [N2O4(g)] = 0.60 M. Calculate the equilibrium concentration for NH3.

Chemistry

1 answer:

natima [27]3 years ago

6 0

The question is incomplete, here is the complete question:

Consider the reaction $4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initially, $[NH_3(g)]=[O_2(g)]$ = 3.60 M; at equilibrium $[N_2O_4(g)]=0.60M$ . Calculate the equilibrium concentration for $NH_3$

Answer: The equilibrium concentration of ammonia is 2.8 M

Explanation:

We are given:

Initial concentration of $[NH_3(g)]$ = 3.60 M

Initial concentration of $[O_2(g)]$ = 3.60 M

For the given chemical equation:

$4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initial: 3.60 3.60

At eqllm: 3.60-4x 3.60-7x 2x 6x

We are given:

Equilibrium concentration of $[N_2O_4(g)]$ = 0.60 M

Evaluating the value of 'x'

$\Rightarrow 2x=0.60\\\\\Rightarrow x=\frac{0.60}{2}=0.2$

So, equilibrium concentration of $NH_3=(3.60-4x)=(3.60-(4\times 0.2))=2.8M$

Hence, the equilibrium concentration of ammonia is 2.8 M

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Answer:

i) 5.00 mL of 1.00 M NaOH: before the equivalence point

ii) 50.0 mL of 1.00 M NaOH: before the equivalence point

iii) 100 mL of 1.00 M NaOH: at the equivalence point

iv) 150 mL of 1.00 M NaOH: after the equivalence point

v) 200 mL of 1.00 M NaOH: after the equivalence point

Explanation:

1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.

Equation:

Molarity = numbrer of moles of solute / volume of the solution in liters.

Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.

Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:

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2. Now calculate the number of moles of NaOH for every condition (addition)

i) 5.00 mL of 1.00 M NaOH

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Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

ii) 50.0 mL of 1.00 M NaOH

$n=0.0500liter\times 1.00M=0.0500molNaOH$

Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

iii) 100 mL of 1.00 M NaOH

$n=0.100liter\times 1.00M=0.100molNaOH$

Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.

iv) 150 mL of 1.00 M NaOH

$n=0.150liter\times 1.00M=0.150molNaOH$

Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.

v) 200 mL of 1.00 M NaOH

This is more volume of NaOH, then this is also after the equivalence point.

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