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Vlada [557]
3 years ago
11

The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2

(M) H2 (M) Initial Rate (M/min.) 0.10 0.10 0.0021 0.10 0.20 0.0084 0.20 0.40 0.0672 What is the unit for the rate constant in the rate law for the formation of ammonia? Choose ONLY ONE and indicate your answer for this question on the exam answer sheet. a. M/min. b. 1/min
Chemistry
1 answer:
notsponge [240]3 years ago
7 0

Answer :  The unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

N_2+3H_2\rightarrow 2NH_3

Rate law expression for the reaction:

\text{Rate}=k[N_2]^a[H_2]^b

where,

a = order with respect to N_2

b = order with respect to H_2

Expression for rate law for first observation:

0.0021=k(0.10)^a(0.10)^b ....(1)

Expression for rate law for second observation:

0.0084=k(0.10)^a(0.20)^b ....(2)

Expression for rate law for third observation:

0.0672=k(0.20)^a(0.40)^b ....(3)

Dividing 2 by 1, we get:

\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2

Dividing 3 by 1 and also put value of b, we get:

\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[N_2]^a[H_2]^b

\text{Rate}=k[N_2]^1[H_2]^2

Now, calculating the value of 'k' by using any expression.

0.0021=k(0.10)^1(0.10)^2

k=2.1M^{-2}min^{-1}

The value of the rate constant 'k' for this reaction is 2.1M^{-2}min^{-1}

That means, the unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

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Explanation:

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Molecular formula of X =?

Next, we shall determine the empirical formula of compound X. This can be obtained as follow:

P = 14.87%

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Divide by the smallest

P = 0.480 / 0.480 = 1

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Empirical formula of compound X is PCl₅

Finally, we shall determine the molecular formula of compound X. This can be obtained as follow:

Molar mass of compound X = 416.48 g/mol

Empirical formula = PCl₅

Molecular formula =?

Molecular formula= [Empirical formula]ₙ

[PCl₅]ₙ = 416.48

[31 + (35.5 × 5)]ₙ = 416.48

[31 + 177.5]n = 416.48

208.5n = 416.48

Divide both side by 208.5

n = 416.48 / 208.5

n = 2

Molecular formula = [PCl₅]ₙ

Molecular formula = [PCl₅]₂

Molecular formula = P₂Cl₁₀

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Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{vap}}{T_b}

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Option C is not true because ions can also carry negative charges and they are called anions.

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