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Alex777 [14]
3 years ago
9

For a certain reaction, Kc = 2.76×103 and kf= 6.54×10−4 M−2⋅s−1 . Calculate the value of the reverse rate constant, kr, given th

at the reverse reaction is of the same molecularity as the forward reaction.Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M−2⋅s−1 include ⋅ (multiplication dot) between each measurement.
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer:

2.37x10⁻⁷ M⁻².s⁻¹

Explanation:

For a generic reversible reaction:

A + B ⇄ C + D

Kf is the constant of the formation of the products (C and D), Kr is the constant of the formation of the reactants (A and B), and Kc is the general equilibrium constant, which is:

Kc = Kf/Kr

2.76x10³ = 6.54x10⁻⁴/Kr

Kr = 6.54x10⁻⁴/2.76x10³

Kr =  2.37x10⁻⁷ M⁻².s⁻¹

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O

Explanation:

O = 3.44

Br = 2.96

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Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this
maw [93]

Answer:  d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g) \Delta H=+1411.1kJ

Reversing the reaction, changes the sign of \Delta H

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)

\Delta H=-1411.1kJ

On multiplying the reaction by \frac{1}{2} , enthalpy gets half:

0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol

Thus the enthalpy change for the given reaction is -705.55kJ

7 0
3 years ago
Which term is dependent on temperature?
Romashka-Z-Leto [24]
1. E. None of the Above
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There are 3 naturally occurring isotopes of hydrogen and 2
DedPeter [7]

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True

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3 0
3 years ago
. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar m
stealth61 [152]

Answer:

<h2>         162g/mol</h2>

Explanation:

The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:

<em>Nicotine has the formula   </em>C_xH_yN_z<em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>

  • <em>1.0 mol of CO₂</em>
  • <em>0.70 mol of H₂O</em>
  • <em>0.20 mol of NO₂</em>

<em>Assume that all the atoms in nicotine are present as products </em>

<h2>Solution</h2>

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

  • 1.0 mol of CO₂ has 1.0 mol of C
  • 0.70 mol of H₂O has 1.4 mol of H
  • 0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

  • C: 1.0
  • H: 1.4
  • N: 0.20

Divide all by the smallest number: 0.20

  • C: 1.0 / 0.20 = 5
  • H: 1.4 / 0.20 = 7
  • N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

  • C:  5mol  × 12g/mol = 60g
  • H: 7mol × 1g/mol = 7 g
  • N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.

5 0
3 years ago
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