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Equilibrium Test - 65

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Equilibrium Test - 65
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Weekly Quiz Competition
  • Question 1
    1 / -0
    For the reaction $$PCl_{5}(g) \longrightarrow PCl_{3}(g)+Cl_{2}(g)$$

     The forward reaction at constant tempearture is favoured by ___________________.
    Solution

  • Question 2
    1 / -0
    The total number of different kind of buffer obtained during the titration of $${ H }_{ 3 }{ PO }_{ 4 }$$ with NaOH are:- 
    Solution

  • Question 3
    1 / -0
    Select the correct acid-base equilibrium.
    Solution

  • Question 4
    1 / -0
    $$HCOOH$$ and $$CH_3COOH$$ solutions have equal $$pH$$. If $$\cfrac{K_1}{K_2}$$ (ratio of ionisation constants of acids) is 4, their molar concentration ratio will be:
  • Question 5
    1 / -0
    At $$90 ^ { \circ } C ,$$ pure water has $$\left[ \mathrm { H } _ { 3 } \mathrm { O } ^ { + } \right] = 10 ^ { - 6 } \mathrm { mol }$$ litre $$^ { - 1 } .$$ The value of $$K _ { w }$$ at $$90 ^ { \circ } \mathrm { C }$$ is:
    Solution

  • Question 6
    1 / -0
    Which of the following salt solution will act as a buffer?
  • Question 7
    1 / -0
    What will be the effect on the equilibrium constant on increasing temperature, if the reaction neither absorbs heat nor releases heat?
    Solution

  • Question 8
    1 / -0
    Hydrogen ion concentration of an aqueous solution is $$1 \times 10^{-4}M$$. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol $$dm^{-3}$$ is __________.
    Solution
    $$\text{Concentration of aqueous solution=} 10^{−4}M$$
    $$\text{If we add equal volume, let say V of water. The final volume of solution will be 2V. The new concentration is}$$
    $$⇒M_1​V_1​=M_2​V_2​$$

    $$⇒10^{−4}M×V=M_2​×2V$$

    $$⇒M_2​=5×10−5M ⟶ \text{New concentration of }[H^+] .$$

    $$pH=−log[H^+]=−log[5×10^{−5}]$$

    $$⇒pH=4.3$$

    $$⇒pOH=14−pH=9.69$$

    $$⇒−log[OH^-]=9.69$$

    $$⇒[OH^-]=10^{−9.69}=2×10^{−10}$$

    $$\text{Thus the concentration of }[OH^-] \ is\  2×10^{−10}M .$$
  • Question 9
    1 / -0
    The dissociation constants for acetic acid  and HCN at $$25^{0}$$C are $$1 . 5 \times 10^{-5}$$  and $$4 . 5 \times 10 ^{-10} $$, respectively. The equilibrium constant for the equilibrium $$CN^{-} + CH_{3}COOH \leftrightharpoons HCN + CH_{3}COO^{-}$$ would be :
    Solution

  • Question 10
    1 / -0
    $$N_{2} + 3 H_{2} \leftrightharpoons 2NH_{3} $$ $$\quad \Delta H= - 21 . 9 Kcal $$

    The favourable conditions for given equilibrium are:
    Solution

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