Question:

Comment on the thermodynamic stability of NO(g) given (1/2)N₂(g)+(1/2)O₂(g)→NO(g);△rH^⊖=90KJMol⁽⁻¹⁾ NO(g)+(1/2)O2(g)→NO2(g):△rH^⊖=−74KJMol⁽⁻¹⁾

Answer:

1. The thermodynamic stability of NO(g) can be determined by looking at the values of the enthalpy change (ΔH^⊖) for the two reactions.

2. The reaction (1/2)N₂(g)+(1/2)O₂(g)→NO(g) has a positive enthalpy change (ΔH^⊖=90kJ/mol), which indicates that the reaction is endergonic and thermodynamically unfavorable.

3. The reaction NO(g)+(1/2)O₂(g)→NO₂(g) has a negative enthalpy change (ΔH^⊖=−74kJ/mol), which indicates that the reaction is exergonic and thermodynamically favorable.

4. Therefore, NO(g) is thermodynamically unstable, as the formation of NO₂(g) is more favorable than the formation of NO(g).

Question:

A reaction A+B→ C+D is found to have a positive entropy change. The reaction will be : A : possible at high temperature B : possible only at low temperature C : not possible at any temperature D : possible at any temperature

Answer:

A : possible at high temperature B : possible only at low temperature C : not possible at any temperature D : possible at any temperature

The reaction A+B→ C+D will be exothermic (i.e. it will release energy) and have a positive entropy change, since the reactants (A and B) will have higher entropy than the products (C and D). This means that the reaction will be favored at high temperatures, since the reactants (A and B) will be more stable at higher temperatures.

Question:

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at −10.0°C △fusH=6.03 KJ mol⁽⁻¹⁾ at 0°C Cp[H2O(1)]=75.3Jmol⁽⁻¹⁾K⁽⁻¹⁾ Cp[H2O(s)]=36.8Jmol⁽⁻¹⁾K⁽⁻¹⁾

Answer:

Answer: Step 1: Calculate the enthalpy change of melting at 0°C: △fusH = 6.03 kJ mol^-1 at 0°C

Step 2: Calculate the enthalpy change of heating 1.0 mol of water from 10°C to 0°C:

△H = Cp x m x ΔT

△H = 75.3J mol^-1 K^-1 x 1.0 mol x (10.0°C - 0°C)

△H = 753 J mol^-1

Step 3: Calculate the enthalpy change of cooling 1.0 mol of water from 0°C to -10°C:

△H = Cp x m x ΔT

△H = 36.8J mol^-1 K^-1 x 1.0 mol x (0°C - (-10.0°C))

△H = 368 J mol^-1

Step 4: Calculate the enthalpy change of freezing 1.0 mol of water at -10°C to ice at -10°C:

△fusH = 6.03 kJ mol^-1 at 0°C

Step 5: Calculate the total enthalpy change on freezing of 1.0 mol of water at 10°C to ice at -10°C:

Total △H = △fusH + △H(heating) + △H(cooling)

Total △H = 6.03 kJ mol^-1 + 753 J mol^-1 + 368 J mol^-1

Total △H = 6.15 kJ mol^-1

Question:

The reaction of cyanamide NH2CN(s) with dioxygen was carried out in a bomb calorimeter and △U was found to be −742.7kJ mol⁽⁻¹⁾ at 298 K. NH²CN(g)+(3/2)O²(g)→N²(g)+CO²(g)+H²O(l) Calculate enthalpy change for the reaction at 298 K?

Answer:

Step 1: Calculate the enthalpy change of the reaction using the given information. △H = -742.7 kJ mol^(-1)

Step 2: Calculate the enthalpy change of the reaction at 298 K. △H = -742.7 kJ mol^(-1) at 298 K

Question:

For the reaction, 2Cl(g)→Cl₂(g) what are the signs of △H and △S ?

Answer:

Answer:

For the reaction, 2Cl(g)→Cl₂(g), the sign of △H is negative (exothermic) and the sign of △S is positive (entropy increases).

Question:

For the reaction at 298 K, 2A+B →C △ H = 400 KJ mol⁽⁻¹⁾ and △ S = 0.02 KJ K⁽⁻¹⁾ mol⁽⁻¹⁾ At what temperature will the reaction become spontaneous considering △ H and △ S to be constant over the temperature range?

Answer:

1. Calculate the change in Gibbs Free Energy (△G) using the equation △G = △H - T△S

2. Set △G = 0 and solve for T.

3. The temperature at which the reaction will become spontaneous is T = △H/△S = 400 KJ mol^(-1)/0.02 KJ K^(-1) mol^(-1) = 20,000 K.

Question:

Calculate the entropy change in surrounding when 1.00 mol of H₂O(l) is formed under standard condition △fH⊖=−286KJmol⁽⁻¹⁾.

Answer:

1. Calculate the amount of heat released when 1.00 mol of H₂O(l) is formed under standard condition.

Q = -286 kJ/mol

2. Calculate the entropy change in surrounding using the equation:

ΔS = -Q/T

where T is the temperature at which the reaction occurs.

For standard conditions, T = 298 K.

Therefore, ΔS = -286 kJ/mol / 298 K = -0.96 kJ/K mol

Question:

The enthalpy of combustion of methane graphite and dihydrogen at 298K are −890.3 kJ mol⁽⁻¹⁾ and −285.8 kJ mol⁽⁻¹⁾ respectively. Enthalpy of formation of CH₄(g) will be: A : −78.8 kJ mol⁽⁻¹⁾ B : −52.27 kJ mol⁽⁻¹⁾ C : +74.8 kJ mol⁽⁻¹⁾ D : +52.26 kJ mol⁽⁻¹⁾

Answer:

1. Enthalpy of formation of a substance is defined as the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states.

2. The enthalpy of combustion of methane, graphite and dihydrogen at 298K are −890.3 kJ mol⁽⁻¹⁾ and −285.8 kJ mol⁽⁻¹⁾ respectively.

3. The enthalpy of formation of CH₄(g) can be calculated using the following equation:

Enthalpy of formation of CH₄(g) = Enthalpy of combustion of methane - Enthalpy of combustion of graphite - Enthalpy of combustion of dihydrogen

4. Substituting the values in the equation:

Enthalpy of formation of CH₄(g) = -890.3 kJ mol⁽⁻¹⁾ - (-285.8 kJ mol⁽⁻¹⁾) - (-285.8 kJ mol⁽⁻¹⁾)

5. Simplifying the equation:

Enthalpy of formation of CH₄(g) = -78.8 kJ mol⁽⁻¹⁾

Therefore, the correct answer is option A: −78.8 kJ mol⁽⁻¹⁾.

Question:

Choose the correct answer. A thermodynamic state function is a quantity :
A : used to determine heat chages B : whose value is independent of path C : used to determine pressure volume work D : whose value depends on temperature only

Answer:

Answer: B : whose value is independent of path

Question:

Calculate the enthalpy change for the process CCl₄(g)→C(g)+4Cl(g) and calculate bond enthalpy of C-Cl in CCl4(g) ΔvapH^⊖(CCl₄)=30.5 kJ mol⁽⁻¹⁾.
ΔfH⊖(CCl₄)=−135.5 kJ mol⁽⁻¹⁾.
ΔaH^⊖(C)=715.0 kJ mol⁽⁻¹⁾.
where ΔaH^⊖ is enthalpy of atomisation
ΔaH^⊖(Cl₂)=242 kJ mol⁽⁻¹⁾.

Answer:

Step 1: Calculate the enthalpy change for the process

ΔH = ΔfH⊖(CCl₄) - [ΔaH^⊖(C) + 4ΔaH^⊖(Cl2)]

ΔH = -135.5 kJ mol^(-1) - [715.0 kJ mol^(-1) + 4(242 kJ mol^(-1))]

ΔH = -135.5 kJ mol^(-1) - [1778 kJ mol^(-1)]

ΔH = -1913.5 kJ mol^(-1)

Step 2: Calculate bond enthalpy of C-Cl in CCl₄(g)

Bond enthalpy of C-Cl = ΔvapH^⊖(CCl₄) - 4ΔaH^⊖(Cl₂)

Bond enthalpy of C-Cl = 30.5 kJ mol^(-1) - 4(242 kJ mol^(-1))

Bond enthalpy of C-Cl = -717 kJ mol^(-1)

Question:

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Answer:

Answer:

1. Identify the given values: Heat absorbed = 701 J Work done = 394 J

2. Calculate the change in internal energy: Change in internal energy = Heat absorbed - Work done Change in internal energy = 701 J - 394 J Change in internal energy = 307 J

Question:

For the reaction, 2 A(g)+ B(g) → 2D(g), △U^⊖=−10.5KJ and △S^⊖=−44.1JK⁽⁻¹⁾ Calculate △G^⊖ for the reaction and predict whether the reaction may occur spontaneously.

Answer:

Step 1: Calculate △G^⊖ using the equation △G^⊖=△U^⊖−T△S^⊖.

Step 2: Substitute the given values for △U^⊖ and △S^⊖ into the equation and calculate △G^⊖.

Step 3: △G^⊖ = -10,500 - (298)(-44.1) = -7,824.8 KJ

Step 4: Compare △G^⊖ to 0. If △G^⊖ is negative, the reaction may occur spontaneously.

Step 5: Since △G^⊖ is negative, the reaction may occur spontaneously.

Question:

For the process to occur under adiabatic conditions, the correct condition is : A : △T=0 B : △p=0 C : q=0 D : w=0

Answer:

Answer: C : q=0

Question:

Enthalpy of combustion of carbon to CO² is -393.5 kJ mol⁽⁻¹⁾. Calculate the heat released upon formation of 35.2g of CO² from carbon and dioxygen gas.

Answer:

Step 1: Calculate the moles of CO² formed from 35.2 g of carbon and dioxygen gas.

Moles of CO² = 35.2 g / (12.011g/mol) = 2.939 mol

Step 2: Calculate the heat released upon formation of 2.939 mol of CO².

Heat released = (2.939 mol) × (-393.5 kJ/mol) = -1150.3 kJ

Question:

Enthalpies of formation of CO(g), CO²(g) N²O(g) and N²O⁴(g) are −110,−393.81 and 9.7kJ mol⁽⁻¹⁾ respectively. Find the value of △rH for the reaction : N²O⁴(g)+3CO(g)→N²O(g)+3CO²(g)

Answer:

Step 1: Find the enthalpies of formation of the reactants and products: CO(g): -110 kJ/mol CO²(g): -393.81 kJ/mol N²O(g): 9.7 kJ/mol N²O⁴(g): 9.7 kJ/mol

Step 2: Calculate the enthalpy change for the reaction: ΔH = [enthalpy of products] - [enthalpy of reactants] ΔH = [(9.7 kJ/mol x 1 mol) + (-393.81 kJ/mol x 3 mol)] - [(9.7 kJ/mol x 1 mol) + (-110 kJ/mol x 3 mol)] ΔH = -1058.43 kJ/mol

Step 3: Find the value of ΔH for the reaction: ΔH = -1058.43 kJ/mol

Question:

The equilibrium constant for a reaction is 10. what will be value of △G^⊖ ? R =8.314 JK−1mol⁽⁻¹⁾, T=300 K

Answer:

1. Calculate the standard free energy change (△G^⊖):

△G^⊖ = -RTlnK

2. Substitute the given values:

△G^⊖ = -(8.314 JK−1mol⁽⁻¹⁾)(300 K)ln(10)

3. Calculate the result:

△G^⊖ = -2494.2 Jmol^(-1)

Question:

N2(g)+3H2(g)→2NH3(g): △rH⊖=−92.4 kJ mol⁽⁻¹⁾ What is the standard enthalpy of formation of NH3 gas?

Answer:

Step 1: The standard enthalpy of formation (ΔHf) is the heat change that occurs when one mole of a compound is formed from its elements in their standard states.

Step 2: The standard enthalpy of reaction (ΔrH⊖) is the heat change that occurs when a reaction occurs between the reactants and products in their standard states.

Step 3: The standard enthalpy of formation of NH3 gas can be calculated using the following equation: ΔHf(NH3) = -ΔrH⊖ + (2 x ΔHf(N2) + 3 x ΔHf(H2)).

Step 4: The standard enthalpy of formation of N2 and H2 gas is 0 since they are in their standard states.

Step 5: Substituting the values into the equation, we get: ΔHf(NH₃) = -92.4 kJ mol^(-1) + (2 x 0 + 3 x 0) = -92.4 kJ mol^(-1).

Therefore, the standard enthalpy of formation of NH3 gas is -92.4 kJ mol^(-1).

Question:

The enthalpies of all elements in their standard states are :
A : unity B : zero C : <0 D : different for each element

Answer:

1. Enthalpy is a measure of the energy of a system.

2. The enthalpy of an element in its standard state is a measure of the energy required to change the element from its standard state to a different state.

3. The enthalpy of element A in its standard state is unity, meaning it requires no energy to change it from its standard state.

4. The enthalpy of element B in its standard state is zero, meaning it requires no energy to change it from its standard state.

5. The enthalpy of element C in its standard state is less than zero, meaning it requires energy to change it from its standard state.

6. The enthalpy of element D in its standard state is different for each element, meaning the energy required to change it from its standard state will vary depending on the element.

Question:

ΔUo of combustion of methane is −X kJ mole⁽⁻¹⁾. The value of ΔHo is: A : =ΔU0 B : >ΔU0 C : <ΔU0 D : =0

Answer:

Answer: C : <ΔU0

Question:

For an isolated system △ U=0 what will be △S?

Answer:

Answer:

For an isolated system, △U=0, then the change in entropy (△S) can be calculated using the equation △S= △Q/T, where △Q is the heat exchange and T is the temperature of the system.

Question:

The enthalpies of all elements in their standard states are:

Answer:

1. Determine the enthalpy of each element in its standard state.

Enthalpy is a measure of the amount of energy stored in a substance. The enthalpy of an element in its standard state is the amount of energy it takes to change the element from its standard state to a gas. Different elements have different enthalpies, and these values can be found in tables of thermodynamic data.

Question:

For an isolated system △ U=0 what will be △S?

Answer:

Answer: ΔS = 0.

The equation ΔU = 0 states that the total internal energy of an isolated system is constant, meaning that the energy of the system cannot change. Therefore, the entropy of the system must also remain constant, and ΔS = 0.