Chemistry Solved SSC-II - Pre-Board Exam - Model Paper - Guess Paper FBISE Class 10

FBISE Class 10 

Chemistry Pre-board Exam - Test Paper and Guess Paper for Exam Preparation 2024

Get ready for your FBISE Class 10 Chemistry exam with this comprehensive pre-board exam test paper and guess paper tailored for the 2024 examination. Prepare effectively and boost your confidence with this curated resource.

Chemistry SSC-II

Section-A (Marks-5)

Q1.Circle the correct option A/B/C/D each part carry equal marks

i-what is the colour of phenphthalein in acidc medium

a)Red b)Orange c)colourless d)pink

i) c) colourless

ii-Formic acid contains a functional group

a) -OH b) -COOH c) –CHO d) –CO

ii) b) -COOH

iii-in which of the following compounds oxygen is attached to two

alky groups

a)alcohol b)alkane c)ether d)phenols

iii) c) ether

iv-Which of the following is a homocyclic compound

a)benzene b)furan c)pyridine d)Thiophene

iv) a) benzene

v-Which of the following method is a way to recognize equilibrium

a)distillation b)halogention c)cracking d)Titration.

v) d) Titration

Section B (marks 15)

Q2.Short questions

i-what are importance of equilibrium?

Equilibrium is a crucial concept in various scientific disciplines, including chemistry, physics, and economics. Its importance lies in several key aspects:

1. Understanding Reaction Dynamics: Equilibrium helps in understanding how chemical reactions proceed and reach a balance between reactants and products. It allows scientists to predict reaction outcomes and design processes with desired yields.

2. Optimization of Industrial Processes: Many industrial processes operate at equilibrium conditions or are controlled to achieve equilibrium. Understanding equilibrium allows engineers and scientists to optimize reaction conditions, such as temperature, pressure, and concentrations, to maximize production efficiency and yield.

3. Environmental Applications: Equilibrium principles are crucial in understanding environmental processes such as the dissolution of gases in water, buffering of pH in natural systems, and chemical equilibrium in the atmosphere. This understanding aids in addressing environmental issues and designing solutions for pollution control and remediation.

4. Chemical Analysis: Equilibrium is fundamental in various analytical techniques, including spectrophotometry, chromatography, and titration. These techniques rely on establishing equilibrium conditions to accurately determine concentrations of substances in samples.

5. Biological Systems: Equilibrium concepts are applicable in biological systems, such as enzyme-substrate interactions, metabolic pathways, and the transport of gases in the bloodstream. Understanding equilibrium in biological processes is vital for advancing medical research and drug development.

6. Thermodynamic Studies: Equilibrium is central to thermodynamics, where it helps in analyzing energy transformations and predicting the spontaneity and direction of chemical reactions. Equilibrium constants and Gibbs free energy changes provide valuable insights into the feasibility of reactions.

7. Predictive Modeling: Equilibrium principles enable scientists and engineers to develop mathematical models to predict the behavior of complex systems. These models are used in fields ranging from chemical engineering and materials science to economics and ecology.

Overall, equilibrium is essential for understanding and manipulating a wide range of natural and artificial processes, making it a fundamental concept in scientific and industrial endeavors.

ii-Why ammonia acts as a Bronsted-lowry base? Give example

Ammonia (NH3) acts as a Bronsted-Lowry base because it can accept a proton (H⁺ ion) from another substance, thus forming its conjugate acid NH4⁺.

In the Bronsted-Lowry theory, acids are substances that donate protons, while bases are substances that accept protons. When ammonia reacts with a substance that can donate a proton, it accepts the proton to form its conjugate acid NH4⁺.

Here's an example of the reaction of ammonia acting as a Bronsted-Lowry base:

NH3 (ammonia) + HCl (hydrochloric acid) → NH4⁺ (ammonium ion) + Cl⁻ (chloride ion)

In this reaction, ammonia (NH3) accepts a proton from hydrochloric acid (HCl) to form ammonium ion (NH4⁺), making ammonia a Bronsted-Lowry base.

iii-State Lewis acid base theory also give examples

The Lewis acid-base theory defines acids as substances that can accept a pair of electrons, and bases as substances that can donate a pair of electrons. According to this theory, the formation of coordinate covalent bonds is the basis for acid-base reactions.

Examples of Lewis acid-base reactions:

1. Formation of Coordination Complexes: Transition metal ions often act as Lewis acids by accepting pairs of electrons from Lewis bases (ligands) to form coordination complexes. For example: ${\text{Ag}}^{+}+2{\text{NH}}_3\to {\text{Ag(NH}}_3{)}_2^{+}$

   In this reaction, the silver ion (${\text{Ag}}^{+}$) acts as the Lewis acid by accepting a pair of electrons from two ammonia molecules (${\text{NH}}_3$) which act as Lewis bases, forming the complex ion ${\text{Ag(NH}}_3{)}_2^{+}$.

2. Formation of Adducts: Lewis acids can form adducts with Lewis bases by accepting electron pairs. For example: ${\text{BF}}_3+{\text{NH}}_3\to {\text{BF}}_3{\text{NH}}_3$

   In this reaction, boron trifluoride (${\text{BF}}_3$) acts as the Lewis acid by accepting a pair of electrons from ammonia (${\text{NH}}_3$), forming the adduct ${\text{BF}}_3{\text{NH}}_3$.

3. Acid-Base Reactions Involving Carbonyl Compounds: Carbonyl compounds such as ketones and aldehydes can act as Lewis acids by accepting pairs of electrons from Lewis bases. For example: ${\text{AlCl}}_3+{\text{CH}}_3\text{COOH}\to {\text{CH}}_3\text{COOH}+{\text{AlCl}}_3$

   In this reaction, aluminum trichloride (${\text{AlCl}}_3$) acts as the Lewis acid by accepting a pair of electrons from acetic acid (${\text{CH}}_3\text{COOH}$), forming a complex between the Lewis acid and base.

These examples demonstrate the versatility of the Lewis acid-base theory in explaining a wide range of chemical reactions involving electron pair donation and acceptance.

iv-Derive unit of Kc for following equilibrium reactions

a) CO+2H2= CH3OH 

b)N2O4=2NO2

To derive the units of ${�}_{�}$ (equilibrium constant) for the given reactions, we first need to understand the definition of ${�}_{�}$. In general, for a reaction:

$��+��\rightleftharpoons ��+��$

The equilibrium constant ${�}_{�}$ is defined as:

${�}_{�}=\frac{[�{]}^{�}[�{]}^{�}}{[�{]}^{�}[�{]}^{�}}$

Where:

• $[�],[�],[�],[�]$ are the molar concentrations of reactants and products at equilibrium, respectively.
• $�,�,�,�$ are the stoichiometric coefficients of the reactants and products in the balanced chemical equation.

Now, let's derive the units of ${�}_{�}$ for the given reactions:

a) $\text{CO}+2{\text{H}}_2\rightleftharpoons {\text{CH}}_3\text{OH}$

The balanced equation indicates that:

• $�=1$ (stoichiometric coefficient of CO)
• $�=2$ (stoichiometric coefficient of H2)
• $�=1$ (stoichiometric coefficient of CH3OH)

Therefore, for reaction (a), the unit of ${�}_{�}$ would be:

${�}_{�}=\frac{[{\text{CH}}_3\text{OH}]}{[\text{CO}][{\text{H}}_2{]}^2}$

The units of ${�}_{�}$ would be (molarity of CH3OH) / (molarity of CO * molarity of H2)².

b) $$N2O4⇌2NO2

The balanced equation indicates that:

• $�=1$ (stoichiometric coefficient of N2O4)
• $�=2$ (stoichiometric coefficient of NO2)

Therefore, for reaction (b), the unit of ${�}_{�}$ would be:

${�}_{�}=\frac{[{\text{NO}}_2{]}^2}{[{\text{N}}_2{\text{O}}_4]}$

The units of ${�}_{�}$ would be (molarity of NO2)² / (molarity of N2O4).

v-Sodium hydroxide solution is use in drain cleaners if concentration of OHin solution of NaOH is 1X10-4M.Is the solution acidic, basic or neutral.

Sodium hydroxide ($\text{NaOH}$) is a strong base, and when dissolved in water, it completely dissociates into sodium ions (${\text{Na}}^{+}$) and hydroxide ions (${\text{OH}}^{-}$). The concentration of hydroxide ions in a solution of $\text{NaOH}$ can be calculated from the concentration of $\text{NaOH}$ itself, assuming complete dissociation.

Given that the concentration of ${\text{OH}}^{-}$ in the solution of $\text{NaOH}$ is $1\times 10^{-4}\text{M}$, we can conclude that the solution is basic.

The solution is basic because the concentration of hydroxide ions (${\text{OH}}^{-}$) is greater than the concentration of hydrogen ions (${\text{H}}^{+}$) in water. In basic solutions, the concentration of ${\text{OH}}^{-}$ is higher than ${\text{H}}^{+}$, resulting in a pH greater than 7.

Section C (marks 10)

Q1.a)Define Law of mass action and derive Kc expression for a

General reaction.[1+3]

Definition of Law of Mass Action: The law of mass action states that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficient in the balanced chemical equation, at a given temperature. It can be represented mathematically as follows:

For the reaction: $��+��\rightleftharpoons ��+��$

The rate of the forward reaction (${\text{R}}_{\text{forward}}$) is proportional to the product of the concentrations of the reactants raised to the power of their respective stoichiometric coefficients ($�$ and $�$):

${\text{R}}_{\text{forward}}\propto [�{]}^{�}\cdot [�{]}^{�}$

Similarly, the rate of the reverse reaction (${\text{R}}_{\text{reverse}}$) is proportional to the product of the concentrations of the products raised to the power of their respective stoichiometric coefficients ($�$ and $�$):

${\text{R}}_{\text{reverse}}\propto [�{]}^{�}\cdot [�{]}^{�}$

Derivation of ${�}_{�}$ Expression: At equilibrium, the rates of the forward and reverse reactions are equal. Therefore:

${\text{R}}_{\text{forward}}={\text{R}}_{\text{reverse}}$

$[�{]}^{�}\cdot [�{]}^{�}=[�{]}^{�}\cdot [�{]}^{�}$

Dividing both sides by the concentrations of the products raised to their stoichiometric coefficients:

$\frac{[�{]}^{�}\cdot [�{]}^{�}}{[�{]}^{�}\cdot [�{]}^{�}}={�}_{�}$

Where ${�}_{�}$ is the equilibrium constant.

b)What is a salt?write any two methods with equations for

prepration of salts[1+2]

Definition of Salt: A salt is a chemical compound composed of positively charged ions (cations) and negatively charged ions (anions) that are held together by ionic bonds. Salts are formed when acids react with bases, and the resulting product is neutralized.

Methods for Preparation of Salts:

1. Acid-Base Neutralization Reaction: When an acid reacts with a base, they neutralize each other, forming water and a salt. For example: $\text{HCl}+\text{NaOH}\to \text{NaCl}+{\text{H}}_2\text{O}$

2. Precipitation Reaction: Salts can also be formed by mixing solutions of two soluble salts. If one of the products formed is insoluble in water, it precipitates out of the solution. For example: ${\text{AgNO}}_3+\text{NaCl}\to \text{AgCl}+{\text{NaNO}}_3$

c)Define and wite one example of Functional group and alkyl

radical[3]

Definition of Functional Group:
A functional group is a specific arrangement of atoms or bonds within a molecule that is responsible for the characteristic chemical reactions of that compound. It imparts particular chemical properties to the molecule. For example, the hydroxyl group (-OH) is a functional group found in alcohols, which gives alcohols their characteristic properties.

Example of Functional Group:
One example of a functional group is the carbonyl group (C=O), found in compounds like ketones and aldehydes. For instance, in acetone (CH3COCH3), the carbonyl group is responsible for its reactivity in various organic reactions.

Definition of Alkyl Radical:
An alkyl radical is a type of functional group composed of carbon and hydrogen atoms derived from an alkane by removing one hydrogen atom. Alkyl radicals are highly reactive and are often involved in various organic reactions.

Example of Alkyl Radical:
An example of an alkyl radical is the methyl radical (${\text{CH}}_3^{\cdot }$), which is derived from methane (${\text{CH}}_4$) by removing one hydrogen atom. Methyl radicals are important intermediates in various radical reactions in organic chemistry.

Tips, Test Paper, and Guess Paper for Preparation

Pre-Board Examination Grade 10th Federal  SSC II

Subject: Chemistry Class 10 

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Exam 2024- Solved - Model Paper

Chemistry Solved SSC-II - Pre-Board Exam - Model Paper - Guess Paper FBISE Class 10

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