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pochemuha
2 years ago
8

The reaction represented above goes essentially to completion. The reaction takes place in a rigid, insulated vessel that is ini

tially at 600 K. 14. What happens to the temperature of the contents 46. What can be inferred about AS for the reaction at of the vessel as the reaction occurs? 600 K? (A) The temperature must increase, because (A) It must be positive, since the reaction is according to Le Châtelier's principle, an increase in temperature causes more products to form. thermodynamically unfavorable at 600 K (B) It must be negative, since there are more moles of products than reactants. C) It must be positive, since AG is negative (B) The temperature must decrease, because the (C) The temperature must decrease, because the (D) The temperature does not change, because the reaction takes place at a temperature above room temperature. and AH is positive. (D) It must be negative, since AG is positive and AH is positive reaction is endothermic. 47. Which of the following statements about the vessel is insulated bonds in the reactants and products is most accurate? . A sample of CH OHg) is placed in the (A) The sum of the bond enthalpies of the bonds previously evacuated vessel with a pressure of P, at 600 K. What is the final pressure in the vessel after the reaction is complete and the contents of the vessel are returned to 600 K? in the reactant is greater than the sum of the bond enthalpies of the bonds in the products (B) The sum of the bond enthalpies of the bonds
Chemistry
1 answer:
Leni [432]2 years ago
5 0

Answer:

44. (C) The temperature must decrease because the reaction is endothermic

45. (D) 3Pi

46. (C) It must be positive since positive since ΔG° is positive and ΔH° is positive

47. (B) The sum of the bond enthalpies of the bonds of the reactant is less than the sum of the bond enthalpies of the bonds of the products

Explanation:

Here we have

CH₃OH(g) → CO(g) + 2H₂(g)   ΔH° +91 kJ/mol_{rxn}

44. Since the reaction is endothermic, absorbs heat, temperature must decrease because the reaction is endothermic

45. Since the number of moles in the reactant is 1 and the number of moles in the product is 3, we have;

Pressure, P is directly proportional to the number of moles

Therefore, where the pressure in the reactant is Pi pressure in the products will be 3Pi

46 Since the reaction takes place spontaneously at 600 K, therefore ΔG is negative and ΔH is positive hence ΔS must be positive

47. Since the reaction is an endothermic reaction, the sum of the bond enthalpies of the bonds of the reactant is less than the sum of the bond enthalpies of the bonds of the products.

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If the temperature increases the molecular movement as well, and if it increases the same it will happen with the molecular movement.

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Explanation:

The pressure is related to the molecular density and the movement that these molecules have.

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3 years ago
The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
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<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 3.45

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

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