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Ludmilka [50]
3 years ago
6

What is Kb for CH3NH2(aq) + H2O(1) CH3NH3(aq) + OH (aq)?

Chemistry
1 answer:
gayaneshka [121]3 years ago
8 0

Answer:

Option B.

Kb = [CH3NH3+] [OH-] / [CH3NH2]

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CH3NH2(aq) + H2O(l) <=> CH3NH3+(aq) + OH- (aq)?

The expression for the equilibrium constant, Kb is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, the expression for the equilibrium constant, Kb fo th above reaction can be written as follow:

Kb = [CH3NH3+] [OH-] / [CH3NH2]

Option B gives the correct expression for Kb.

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Answer:

Explanation:

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m = 0.5Kg water = 500g water

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3 0
3 years ago
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Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

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Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

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