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mrs_skeptik [129]
2 years ago
5

The isomerization of methyl isonitrile, CH3NC, to acetonitrile, CH3CN, was studied in the gas phase at 215oC, and the following

data were obtained: Time (s) [CH3NC] (M) 0 0.0165 2,000 0.0110 5,000 0.00591 8,000 0.00314 12,000 0.00137 15,000 0.00074 (a) Calculate the average rate of reaction, in M/s, for the time interval between each measurement.
Chemistry
1 answer:
Viefleur [7K]2 years ago
3 0

Answer:

2.75 × 10⁻⁶ M/s

1.69 × 10⁻⁶ M/s

9.23 × 10⁻⁻⁷ M/s

4.43 × 10⁻⁻⁷ M/s

2.1 × 10⁻⁻⁷ M/s

Explanation:

We have the following information for the isomerization of methyl isonitrile

Time (s)      [CH₃NC] (M)

  0              0.0165

2000          0.0110

5000          0.00591

8000          0.00314

12000         0.00137

15000         0.00074

To calculate the average rate of reaction (r) for each interval, we need to use the following expression:

r = -Δ[CH₃NC]/Δt

Interval 0-2000 s

r = - (0.0110 M-0.0165 M)/2000 s - 0 s = 2.75 × 10⁻⁶ M/s

Interval 2000-5000 s

r = - (0.00591 M-0.0110 M)/5000 s - 2000 s = 1.69 × 10⁻⁶ M/s

Interval 5000-8000 s

r = - (0.00314 M-0.00591 M)/8000 s - 5000 s = 9.23 × 10⁻⁻⁷ M/s

Interval 8000-12000 s

r = - (0.00137 M - 0.00314 M)/12000 s - 8000 s = 4.43 × 10⁻⁻⁷ M/s

Interval 12000-15000 s

r = - (0.00074 M - 0.00137 M)/15000 s - 12000 s = 2.1 × 10⁻⁻⁷ M/s

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The correct mathematical expression for finding the molar solubility (s) of calcium phosphate is
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Answer is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.

<span> Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] = 3s(Ca₃(PO₄)₂) = 3s.<span>
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For a principal quantum number n, how many atomic orbitals are possible?
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The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r
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Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

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