Question

9. Consider the exothermic reaction 2 NHs (aq)+ H2COs (aq) s (NH22CO(aq)+2 H2O () a. 1.0 moles of NH3 and 3.0 moles of (NH2)2CO are added to 1.0 L of water. Calculate the initial value of Q before any reaction takes place.

Answer 1

a)

[NH₃] = moles/L = 1.0 mol/1.0 L = 1.0 M

[H₂CO₃] = moles/L = 3.0 mol/1.0 L = 3.0 M

	2NH3(aq)	H2CO3(aq) <==>	(NH2)2CO(aq)	2H2O(l)	Q
Initial	1.0 M	0	3.0 M	-	[NH3] = moles/L = 1.0 mol/1.0 L = 1.0 M [H2CO3] = moles/L = 3.0 mol/1.0 L = 3.0 M Q = [(NH_2)_2 CO]/[NH_3]^2 [H_2 CO_3] but [H2CO3] = 0 M Q = [3. 0]/[1. 0]^2[0] before reaction takes place Q = infinity b) we got in a) that Q = infinity 1) the reaction is spontaneous in back direction False (Q > k) 2) 2) the reaction is non-spontaneous: True (Q > K) 3) when Q = infinity, reaction is at equilibrium: false 4) Reactants predominant at equilibrium: false 5) products will predominant at equilibrium: true (because K > 1, [products] > [reactants]) 6) Reactants and products are in equal proportions at equilibrium: false (because products concentration > reactants) 7) before any reaction occurs products are predominate: true (because Q = infinity) 8) before any reaction occurs reactants are predominate: false c) Finding Q

$Q = \frac{[(NH_2)_2CO]}{[NH_3]^2[H_2CO_3]}$

But [H2CO3] = 0 M

$Q = \frac{[3.0]}{[1.0]^2[0]}$

before reaction takes place, $Q =\infty$ .

b)

We got in a) that $Q =\infty$

1) The reaction is spontaneous in back direcion, False (Q >K)

2) the reaction is non-spontaneous: True (Q>K)

3) when $Q =\infty$ , reaction is at equilbrium : false

4) Reactants predominant at equilbrum: false

5) products will predominant at equilibrium: true (because K > 1, [products]> [reactants])

6) Reactants and products are in equal proportions at equilibrium : false (because products concentration > reactants)

7) before any reaction occurs products are predominate : true (because Q = infinity)

8) before any reaction occurs reactantss are predominate : false

c)

Finding Q

$Q=\frac{[(NH_2)_2CO]}{[NH_3]^2[H_2CO_3]}$

$Q=\frac{[0.1M]}{[1.0M]^2[0.1M]}$

$Q=1$

Q < K

Q/K = 1/1000 = 0.0001

To increase Q/K ratio

1) Add products

By adding products Q will increase because Q = products/ reactants, Q/K will increase

2) Add reactants

Q will decrease so Q/K ratio will decrease

3) Heat the mixture

Given reaction is exothermic. So if heat the mixture then Reactants concentration will increase, k decreases

So Q/K will increase

4) Cool the mixture

Given reaction is exothermic, so the K will increase. Q/K decreases

5) The reaction is not at equilibrium (Q $\neq$ K)

6) Water has no effect