# Temperature will change in a system until this is reached...

## Answers

1) The answer is: Energy must have been absorbed from the surrounding environment.

There are two types of reaction:

1) Endothermic reaction (ΔH > 0, chemical reaction that absorbs more energy than it releases). In endothermic reactions heat is reactant.

Because products have higher energy, this example is endothermic reaction.

2) Exothermic reaction (chemical reaction that releases more energy than it absorbs).

2) The answer is: A reversible reaction contains a forward reaction, which occurs when reactants form products, and a reverse reaction, which occurs when products form reactants.

For example, balanced reversible chemical reaction: N₂ + 3H₂ ⇄ 2NH₃.

Nitrogen and hydrogen are reactants and ammonia is product of reaction. Reaction goes in both direction. Ammonia is synthesized from nitrogen and hydrogen and ammonia decomposes on nitrogen and hydrogen.

The amount of substance of reactants and products of reaction do not change when chemical reaction is in chemical equilibrium.

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which not change with time. Speed of direct and irreversible chemical reaction are equal

3) The answer is: The reaction rate of the forward reaction would increase in order to decrease the number of particles.

According to Le Chatelier's Principle the position of equilibrium moves to counteract the change.

The equilibrium shift to the right, so more product (ammonia) will be produced.

There are less molecules of ammonia than molecules of hydrogen and nitrogen. For every two molecules of ammonia, there are four molecules of hydrogen and nitrogen.

4) The answer is: The increase in energy will cause the reactants' particles to move faster, which will increase their temperature and lead to a faster reaction rate.

The reaction rate is the speed at which reactants are converted into products.

The collision theory states that a certain fraction of the collisions (successful collisions) cause significant chemical change.

The successful collisions must have enough energy (activation energy).

Chemical bonds are broken and new bonds are formed.

Particles are in constant, random motion and possess kinetic energy, molecules faster and have more collisions.

5) The answer is: When a reaction is at chemical equilibrium, a change in the system will cause the system to shift in the direction that will balance the change and help the reaction regain chemical equilibrium.

For example, chemical reaction: heat + NH₄⁺ + OH⁻ ⇄ NH₃ + H₂O.

According to Le Chatelier's Principle, the position of equilibrium moves to counteract the change, because heat is increased, system consume that heat, so equilibrium is shifted to right, by decreasing concentration of reactants and increasing concentration of product.

[tex]\Delta s\ =\ 21.33\ J/K[/tex]

Explanation:

Given,

Mass of the ice = [tex]m_i\ =\ 13.8\ kg\ =\ 0.0138\ kg[/tex]Temperature of the ice = [tex]T_i\ =\ 0^o\ C[/tex]Mass of the original water = [tex]m_w\ =\ 134\ g\ =\ 0.134\ kg[/tex]Temperature of the original water = [tex]T_w\ =\ 70.0^o\ C[/tex]Specific heat of water = [tex]S_w\ =\ 4190\ J/kg K[/tex]Latent heat of fusion of ice = [tex]L_f\ =\ 333\ kJ/kg[/tex]Let T be the final temperature of the mixture,

Therefore From the law of mixing, heat loss by the water is equal to the heat gained by the ice,

[tex]m_iL_f\ +\ m_is_w(T_f\ -\ 0)\ =\ m_ws_w(T_w\ -\ T_f)\\\Rightarrow 333000\times 0.0138\ +\ 0.0138\times 4190T_f\ =\ 0.134\times 4190\times(70.7\ -\ T_f)\\\Rightarrow 4595.4\ +\ 57.96T_f\ =\ 39789.96\ -\ 562.8T_f\\\Rightarrow 620.76T_f\ =\ 35194.56\\\Rightarrow T_f\ =\ \dfrac{35194.56}{620.76}\\\Rightarrow T_f\ =\ 56.69^o\ C[/tex]

Now, We know that,

Change in the entropy,

[tex]\Delta s\ =\ s_f\ -\ s_i\ =\ \dfrac{Q}{T}\\\Rightarrow \displaystyle\int_{s_i}^{s_f} ds\ =\ \displaystyle\int_{T_i}^{T_f}\dfrac{msdT}{T}\\\Rightarrow \Delta s =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]

Now, change in entropy for the ice at 0^o\ C to convert into 0^o\ C water.

[tex]\Delta s_1\ =\ \dfrac{Q}{T}\\\Rightarrow \Delta s_1\ =\ \dfrac{m_iL_f}{T}\ =\ \dfrac{0.0138\times 333000}{273.15}\ =\ 16.82\ J/K.[/tex]

Change in entropy of the water converted from ice from [tex]273.15\ K[/tex] to water 330.11 K water.

From the equation (1),

[tex]\therefore \Delta s_2\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_2\ =\ 0.0138\times 4190\times \ln \left (\dfrac{273.15}{330.11}\ \right )\\\Rightarrow \Delta s_2\ =\ 6.88\ J/K[/tex]

Change in entropy of the original water from the temperature 342.85 K to 330.11 K

From the equation (1),

[tex]\therefore \Delta s_3\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_3\ =\ 0.134\times 4190\times \ln \left (\dfrac{330.11}{343.85} \right )\\\Rightarrow \Delta s_3\ =\ -2.36\ J/K[/tex]

Total entropy change = [tex]\Delta s\ =\ \Delta s_1\ +\ \Delta s_2\ +\ \Delta s_3\\\Rightarrow \Delta s\ =\ (16.82\ +\ 6.88\ -\ 2.36)\ J/K\\\Rightarrow \Delta s\ =\ 21.33\ J/K.[/tex]

Hence, the change in entropy of the system form then untill the system reaches the final temperature is 21.33 J/K

[tex]$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$[/tex]

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K

Explanation:

We are given that carbon dioxide undergoes a process in a closed system.

We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.

The entropy change of the carbon dioxide is given by

[tex]$ S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1}) $[/tex]

Where Cp is the specific heat constant

Cp = 0.846 kJ/kg.K

R is the universal gas constant

R = 0.1889 kJ/kg.K

T₁ and T₂ is the initial and final temperature of carbon dioxide.

P₁ and P₂ is the initial and final pressure of carbon dioxide.

[tex]$ S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50}) $[/tex]

[tex]$ S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888) $[/tex]

[tex]$ S_2 - S_1 = 0.5864 - 0.6968 $[/tex]

[tex]$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$[/tex]

Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K

[tex]\Delta S=-0.11\frac{kJ}{kg*K}[/tex]

Explanation:

Hello,

In this case, we can compute the entropy change by using the following equation containing both pressure and temperature:

[tex]\Delta S=Cp\ ln(\frac{T_2}{T_1} )-R\ ln(\frac{p_2}{p_1} )[/tex]

Thus, we use the given data to obtain (2 MPa = 2000 kPa):

[tex]\Delta S=0.846\frac{kJ}{kg*K} \ ln(\frac{800K}{400K} )-0.1889\frac{kJ}{kg*K} \ ln(\frac{2000kPa}{50kPa} )\\\\\Delta S=-0.11\frac{kJ}{kg*K}[/tex]

Best regards.

Explanation:

Let us take the equilibrium temperature is T.

Now, the heat absorbed by the ice is equal to the heat lost to the water.

Therefore, the formula will be as follows.

[tex]m_{water} \times C_{water} (T - 87.3) + m_{ice} \times L_{f} + m_{ice} \times C_{water} (T - 0) = 0[/tex]

Therefore, putting the given values into the above formula as follows.

[tex]m_{water} \times C_{water} (T - 87.3) + m_{ice} \times L_{f} + m_{ice} \times C_{water} (T - 0) = 0[/tex]

[tex]150 g \times 4190 J/kg K (T - 87.3) + 0.0102 kg \times 333 kJ/kg + m_{ice} \times C_{water} (T - 0) = 0[/tex]

628500T - 54868050 + 3.3966 + 42.738T = 0

T = [tex]\frac{54868046.6}{628542.738}[/tex]

= [tex]87.29^{o}C[/tex]

or, = (87.29 + 273.15)K

= 360.44 K

Since, formula for entropy change of ice is as follows.

[tex]\Delta S_{ice} = \frac{m_{ice} \times L_{f}}{T_{ice}} + m_{ice} C ln \frac{T}{T_{melting}}[/tex]

= [tex]\frac{0.0102 [\frac{333000}{273.15} + 4190 \times ln (\frac{360.44}{273.15})][/tex]

= [tex]0.0102 \times 2381.015[/tex]

= 24.286 J/K

Entropy change for water ([tex]\Delta S_{water}[/tex]) = [tex]m_{ice} \times C \times ln (\frac{T}{T_{initial})}[/tex]

= [tex]0.0102 \times 4190 \times ln \frac{360.44}{360.45}[/tex]

= [tex]-1.186 \times 10^{-3}[/tex] J/K

Therefore, calculate the net change in entropy as follows.

[tex]\Delta S = \Delta S_{water} + \Delta S_{ice}[/tex]

= [tex]-1.186 \times 10^{-3} + 24.286[/tex] J/K

= 24.28481 J/K

Thus, we can conclude that the net entropy change of the system from then until the system reaches the final (equilibrium) temperature is 24.28481 J/K.

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

ΔS = 0.85244 KJ/k

This would be canberra.

Hydrogen is the most abundant element on the periodic table..hope it .