KPK 11th Class Chemistry Chapter 11 Thermochemistry Short Questions Answers
| Class: | 11th Class | Subject: | Chemistry |
| Chapter: | Chapter 11 | Board: | KPK Boards |
KPK 11th Class Chemistry Chapter 11 Thermochemistry Short Questions with answers are combined for all 11th class(Intermediate/hssc) Level students.Here You can prepare all Chemistry Chapter 11 Thermochemistry short question in unique way and also attempt quiz related to this chapter. Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.
Class/Subject: 11th Class Chemistry
Chapter Name: Thermochemistry
Board: All KPK Boards
- Malakand Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
- Mardan Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
- Peshawar Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
- Swat Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
- Dera Ismail Khan Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
- Kohat Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
- Abbottabad Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
- Bannu Board 11th Class Chemistry Chapter 11 Thermochemistry short questions Answer
Helpful For:
- All KPK Boards 11th Class Chemistry Annual Examination
- Schools 11th Class Chemistry December Test
- KPK 11th Class Chemistry Test
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KPK 11th Class Chemistry Chapter 11 Thermochemistry Short Questions Answers
Reversing equation 2 and then solving equation no. 1 and three simultaneously I2(s) I2(g)∆H°=sub=62.3 KJmol-1Applying Hess’s law, calculate ∆H° for the sublimation of one mole of iodine from the following equations. H2(g) +12 I2(s) 2HI(g)∆H°=51.8 kjmol-1 H2(g) + I2(g) 2HI(g)∆H°=-10.5 kjmol-1 I2(s) I(g)∆H°=sub=?
HCL(g)+ aq HCl (aq) ∆H° =-72.41 KJ mol-1 NH3(ag) +HCl (aq) NH4Cl(aq)∆H° =-51.48 KJ mol-1 Solving all the three equation simultaneously. We get NH3(ag)+ HCl(aq) NH4Cl(aq)∆H° =-159.05 kJ mol-1 C2H5OH(1)+ 3O2 → 2CO2(g)+3H2(g)∆ H10 =-1402.14KJ/mole Eq.no.1 C(s)+O2(g) → CO2(g)∆H20 =-393.7KJ/mole Eq.no.2 H2(g) +1/2 O2(g) → H2O(1) ∆H30 =-285.81 KJ/mole Eq.no.3 Calculate the heat of formation of an aqueous solution of NH4Cl from the following data.
Liquid ethanol when burnt in oxygen at 250c shows ∆H° =-1402.14 KJ mol-1. The heat of formation of CO2 and H2O are- 393.50 and- 285.81 KJ mol-1 respectively at the same temperature. Calculate the heat of formation of ethanol at 25°C.
This law is merely the law of conservation of energy i.e. the total amount of energy of the universe is constant however; energy can be transformed from one form to another. Both heat and work are the form of energy. So the change in internal is equal to the sum of its heat changes and work. Thus the first law of thermodynamics can be written mathermatically as: ∆E = q + w Where as ∆E =change in internal energy Q = heat evolved or absorbed W = work doneExplain the following short questions with reasons. Total energy of the system and its surrounding remains constant.
How to Write Perfect Short Answers?
In Board Exams, the examiner looks for specific keywords and presentation. Here is how to attempt Chapter 11 questions:
- Ideal Length: Write 3 to 5 lines for each short question. Too short gets fewer marks, too long wastes time.
- Highlighting: Use a Blue Marker to highlight key dates, names, or scientific terms in your answer.
- Units & Formulas: Always write the formula and SI unit. Without units, 0.5 marks are deducted.
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