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ECAT Chemistry Chapter 8 Chemical Equilibrium Online Test
| Sr. # | Questions | Answers Choice |
|---|---|---|
| 1 | N2 +3H2⇌
2NH3 + Heat for above equation, the |
Low temperature at high pressure. High temperature and low pressure. High temperature and high pressure. Low temperature at low pressure. |
| 2 | The substance which increases rate of reaction but remains unchanged at the end of reaction is called : | Catalyst. Indicator. Promoter. Activator. |
| 3 | N23H2⇌ 2NH3 Which of the following change will favorthe formation of moreNH3at equilibrium in above reaction : |
By adding NH<sub>3.</sub> By removing<span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">H</span><sub>2.</sub> By decreasing pressure. By increasing pressure. |
| 4 | Extent to H2 + L2 à2Hl can be increased by
: |
<p class="MsoNormal">Increasing temperature.<o:p></o:p></p> <p class="MsoNormal">Increasing product. <o:p></o:p></p> <p class="MsoNormal">Increasing pressure. <o:p></o:p></p> <p class="MsoNormal">Adding a catalyst.<o:p></o:p></p> |
| 5 | 2SO2+ O2⇌2SO2H= 188KJ mole-1
Which statement about following equilibrium is correct : |
<p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">The value of</span>K<sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif;">p</span></sub>falls with arise in temperature.<o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">The value of</span>K<sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif;">p</span></sub>is equal tok<sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif;">c.</span><o:p></o:p></sub></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">The value of</span>K<sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif;">p</span></sub>falls with the increase pressure.<o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">Adding V</span><sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif;">2</span></sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">O</span><sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif;">5</span></sub><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">catalyst increase the equilibrium yield of Sulphur trioxide.</span><o:p></o:p></p> |
| 6 | 2SO2 + O2 ⇌ 2SO2 H= 188KJ mole-1 |
The value of<span style="font-size:11.0pt;line-height:107%; font-family:"Calibri",sans-serif;mso-ascii-theme-font:minor-latin;mso-fareast-font-family: Calibri;mso-fareast-theme-font:minor-latin;mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman";mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US;mso-fareast-language:EN-US;mso-bidi-language:AR-SA">K<sub>p</sub>falls with arise in temperature.</span> The value of<span style="font-family: Calibri, sans-serif; font-size: 11pt; line-height: 15.6933px;">K<sub>p</sub>is equal to</span><span style="font-size:11.0pt;line-height:107%; font-family:"Calibri",sans-serif;mso-ascii-theme-font:minor-latin;mso-fareast-font-family: Calibri;mso-fareast-theme-font:minor-latin;mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman";mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US;mso-fareast-language:EN-US;mso-bidi-language:AR-SA">k<sub>c.</sub></span> The value of<span style="font-family: Calibri, sans-serif; font-size: 11pt; line-height: 15.6933px;">K<sub>p</sub>falls with the increase pressure.</span> Adding V<sub>2</sub>O<sub>5</sub> catalyst increase the equilibrium yield of Sulphur trioxide.<p class="MsoNormal"><o:p></o:p></p> |
| 7 | A large value of Kcmeans that at equilibrium : | Less reactant and more products. Reactants and product in same amounts. More reactants and less products. None of above. |
| 8 | In exothermic reversible reaction increase in temperature shift the equilibrium to : | Remains unchanged. Product side. Reactant side. None of above. |
| 9 | N2O4⇌
2NO2 For the above reaction, which of the Following expression of
Kc correct : |
<p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">Kc = [ N<sub>2</sub>O<sub>4</sub>]/[ NO<sub> 2</sub>]<sup>2</sup><o:p></o:p></p> <p class="MsoNormal">Kc = [ N<sub>2</sub>O<sub>4</sub>]/ [NO<sub> 2</sub>]<o:p></o:p></p> <p class="MsoNormal">Kc = [ N<sub>2</sub>O]<sup>2</sup>/[ N<sub>2</sub>O<sub> 4</sub>]<o:p></o:p></p> <p class="MsoNormal">Kc = [ N<sub>2</sub>O<sub>4</sub>]/[ N O<sub> 2</sub>]<o:p></o:p></p> |
| 10 | Ifkcof a reaction productis verylarge, it indicates that equilibrium occurs : | With the help of a catalyst. With no forward reaction. At a low product concentration. At a high product concentration. |
| 11 | Almost forward reaction is complete when value of kc : | Neither larger nor very small. Very small. Very large. Negligible. |
| 12 | In the particular reaction for the valueKc 1 x10-25which statement is correct : | Almost forward reaction is completed. Amount of reactant is negligible as compared to product. Amount of product is negligible as compared to reactant. Amount of product is equal to amount of reactant. |
| 13 | For what value ofKc almost forward reaction is complete : | <p class="MsoNormal">K<sub>c = </sub>10<sup>30</sup><o:p></o:p></p> <p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">K<sub>c = </sub>10<sup>-30</sup><o:p></o:p></p> <p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">K<sub>c = </sub>0<o:p></o:p></p> <p class="MsoNormal">K<sub>c = </sub>1<o:p></o:p></p> |
| 14 | N2 + 3H2 ⇌2NH3 The unit of Kc for tis reaction will be: |
<p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">mol<sup>2</sup> dm<sup>-6</sup></span><o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">mol<sup>-2</sup> dm<sup>+6</sup></span><o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">mol dm<sup>-3</sup></span><o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">mol<sup>-1</sup> dm<sup>+3</sup></span><o:p></o:p></p> |
| 15 | N2 + O2 ⇌2NO The unit of Kc for tis reaction will be: |
<p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">No unit<o:p></o:p></p> <p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">mol<sup>-2</sup> dm<sup>-3</sup></span><o:p></o:p></p> <p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">mol<sup>-1</sup> dm<sup>-3</sup></span><o:p></o:p></p> <p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">mol<sup>-2</sup> dm<sup>+3</sup></span><o:p></o:p></p> |
| 16 | 1 mol of N2O4 was decomposed according
to given equation in 1dm3 container. At equilibrium x mole of N2O4
have dissociated. What is the value of KC: |
<p class="MsoNormal">2x/(1-x)<sup>2<o:p></o:p></sup></p> <p class="MsoNormal">4x<sup>2</sup>/(1-x)<o:p></o:p></p> <p class="MsoNormal">4x/(1-x)<o:p></o:p></p> <p class="MsoNormal">2x/(1-x)<o:p></o:p></p> |
| 17 | The correct relation b/wKc andKp is : | <p class="MsoNormal">K<sub>p</sub> = K<sub>c</sub>[P/N] <sup>Δn<o:p></o:p></sup></p> <p class="MsoNormal">K<sub>c = </sub>K<sub>p</sub>(RT) <sup>Δn</sup><o:p></o:p></p> <p class="MsoNormal">K<sub>p = </sub>K<sub>c</sub>(RT) <sup>Δn</sup><o:p></o:p></p> <p class="MsoNormal">K<sub>p = </sub>K<sub>c</sub>(RT) <sup>Δn</sup><o:p></o:p></p> |
| 18 | For the above reaction the relationship b/w kc and kp will be : | <p class="MsoNormal">K<sub>p</sub> = K<sub>c</sub>RT<o:p></o:p></p> <p class="MsoNormal">Kp = K<sub>c</sub>(RT)<sub>-1</sub><o:p></o:p></p> <p class="MsoNormal">K<sub>p</sub> = K<sub>c</sub>(RT)<sub>-2</sub><o:p></o:p></p> <p class="MsoNormal">K<sub>c</sub> = K<sub>p</sub><o:p></o:p></p> |
| 19 | H2 + L2 ----2Hl In the above equilibrium system, if the concentration of reactants at 25°C is increased, the value KC will : |
Remains Constant Increases Cecreases Depends upon nature of reactans |
| 20 | I a chemical reaction equilibrium is said to have been established when : | Rate of opposing reactions are equal. Rate constants of opposing reactions are equal. Opposing reactions stop. Concentration of reactants and products are equal |
| 21 | Law of mass action was given by : | Guldberg and Waage. Berkeley and Hartly. Ramsay and Reyleigh. Berthelot. |
| 22 | The rate of reaction : | Remain same as reaction proceeds. May decrease or increase as reaction proceeds . Increase as reaction proceeds. Decreases as reaction proceeds. |
| 23 | A chemical reaction A---------->B is said to be in equilibrium when : | Rate of transformation of A to B is equal to B to A. 50% reactant has been changed to B. Conversion of A to B is 50% complete Complete conversion of A to B has taken place. |
| 24 | The rate of a chemical reaction is directly;y proportional to product of molar concentration of reaction substance it is called : | Low of conservation of energy. Law of mass action. Rate law . Active mass rule. |
| 25 | What happens when reaction is at equilibrium and more reactant is added : | Forward reaction rate is increased. Forward reaction rate is decreased. Backward reaction rate is increased. Equilibrium remains unchanged. |
| 26 | A reaction is reversible because : | Products are stable. Reactants are reactive. Products are reactive. Reactants re stable. |
| 27 | An excess of aqueous silver nitrate is added to aqueous
barium chloride and precipitate is removed by filtration. What are the main ion
in filtrate? |
<p class="MsoNormal">Ag<sup>+</sup> and NO<sub>3-</sub> only<o:p></o:p></p> <p class="MsoNormal">Ag<sup>+</sup> and Ba<sup>2</sup> and NO<sup>-3</sup><o:p></o:p></p> <p class="MsoNormal">Ba<sup>2</sup> and NO<sup>-3</sup><o:p></o:p></p> <p class="MsoNormal">Ba<sup>2</sup> and NO<sup>-3 </sup>and Cl-<o:p></o:p></p> |
| 28 | The solubility product of
AgCl is 2.0 x 10-3 mol2 dm-6 , The maximum
concentration of Ag ion in the solution is : |
<p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;"><o:p></o:p></span></p> <p class="MsoNormal"><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">2.0 x 10<sup>-10</sup> mol dm<sup>-3</sup></span><o:p></o:p></p> <p class="MsoNormal">1.41 <span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">x 10<sup>-10</sup> mol dm<sup>-3<o:p></o:p></sup></span></p> <p class="MsoNormal">1.0<sup><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;"> </span></sup><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">x 10<sup>-10</sup> mol dm<sup>-3<o:p></o:p></sup></span></p> <p class="MsoNormal">4.0<sup><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;"> </span></sup><span style="font-size: 10.5pt; line-height: 107%; font-family: Arial, sans-serif; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">x 10<sup>-10</sup> mol dm<sup>-3</sup></span><o:p></o:p></p> |
| 29 | The ph of 10-3 mole dm-3 of an aqueous solution of H2SO4
is : |
<p class="MsoNormal">3.0<o:p></o:p></p> <p class="MsoNormal">2.7<o:p></o:p></p> <span style="font-size:11.0pt;line-height:107%; font-family:"Calibri",sans-serif;mso-ascii-theme-font:minor-latin;mso-fareast-font-family: Calibri;mso-fareast-theme-font:minor-latin;mso-hansi-theme-font:minor-latin; mso-bidi-font-family:Arial;mso-bidi-theme-font:minor-bidi;mso-ansi-language: EN-US;mso-fareast-language:EN-US;mso-bidi-language:AR-SA">2.0</span> <p class="MsoNormal">1.5<o:p></o:p></p> |
| 30 | Which statement about the following equilibrium in correct? 2SO2 (g)+ O2(g)---------------2sO3(g)H= - 188.3 KJ mol-1 |
T value of K<sub>p falls witha rise in temperate.</sub> The value of K<sub>p falls withincreasing pressure</sub><p class="MsoNormal"><o:p></o:p></p> Adding V<sub>2O</sub><sub>5</sub>catalyst increase the equilibrium yield of sulfur trioxide<p class="MsoNormal"><o:p></o:p></p> The value of K<sub>p is equal to</sub>K<sub>p</sub> <br><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal"><o:p></o:p></p> |
