3 Marks Question

Question 513 Marks

A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the inoisation constant of pyridine.

Answer

$\text{C}_6\text{H}_5\text{N}+\text{HCl}^-+\text{H}_2\text{O}\rightleftharpoons\text{C}_6\text{H}_5\text{N}^+\text{HOH}^-+\text{HCl}$
(Solution is acidic due to hydrolysis)
$\text{pH}=7-\frac{\text{pK}_\text{b}}{2}-\frac{\log\text{C}}{2}$
$3.44=7-\frac{\text{pK}_{\text{b}}}{2}-\frac{\log0.02}{2}$
$\text{pK}_{\text{b}}=8.82$
$\text{pK}_{\text{b}}=8.82=-\log\text{ pK}_{\text{b}}$
$\log\text{K}_{\text{b}}=-8.82=\bar{9}.18$
$\text{K}_{\text{b}}=\text{antilog }\bar{9}.18$
$=1.5\times10^{-9}$

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Question 523 Marks

On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$
What will be the effect of addition of argon to the above reaction mixture at constant volume?

Answer

$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$ Effect of temperature: All chemical reactions are accompanied by the liberation or uptake if heat. If we regard heat as a reactant or product in an endothermic or exothermic reaction respectively, we can use the Le Chatelier principle to predict the direction in which an increase or decrease in temperature will shift the equilibrium state. Therefore for the give reaction we can write, $\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$ The Le Chatelier principle tells us that a net reaction will occur in the direction that will partially counteract this change. Since the reaction is exothermic, a shift of the equilibrium to the left will take place. Thus, by decreasing the temperature yield of $NH_3$ can be increased. Effect of pressure:
$\Delta\text{n}_\text{g}=\text{n}_\text{g}\text{(products)}-\text{n}_\text{g}\text{(reactants)}$ $= 2 - (1 + 3) = 2 - 4 = -2$ Since $\Delta\text{n}_\text{g}$ is negative, if the pressure is increased,The equilibrium will shift in the forward direction according to the Lr Chatelier principle, because doing this will decrease the total number of moles of gases and hence the pressure which will decrease the effect of increasing pressure. Thus, high pressure would increase the yield of ammonia. Addition of argon at constant volume The addition of argon at constant volume would not change the partial pressures of any substance, $N_2, H_2$ or $NH_3$. Hence, the equilibrium is not disturbed and there would not be any effect of addition of argon at constant Volume.

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Question 533 Marks

$2\text{NO}(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g}),\Delta\text{H}=-117\text{J}$
Predict the effect of an increase in concentration of NO on the equilibrium concentration of $\text{NO}_2$.
Predict the effect of pressure decrease as a result of increased volume on the equilibrium concentration of $\text{NO}_2$.

Answer

$2\text{NO}(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\Delta\text{H}=-117\text{J}$

If we increase the concentration of $\text{NO}_2$ the rate of forward reaction will increase, i.e. more $\text{NO}_2$ will be formed.
Decrease in pressure will favour backward reaction, i.e. less $\text{NO}_2$ will be formed because number of moles of reactants are more than products.

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Question 543 Marks

$\begin{matrix}\text{A }\rightleftharpoons\text{ B }&\text{K}_1=1&\text{B }\rightleftharpoons\text{ C}&\text{K}_2=2\\\text{C }\rightleftharpoons\text{ D}&\text{K}_3=3&\text{D }\rightleftharpoons\text{ E}&\text{K}_4=4\end{matrix}$
What is value of K for $\text{A }\rightleftharpoons\text{ E}?$

Answer

$\text{A }\rightleftharpoons\text{ E K}=\frac{[\text{E}]}{[\text{A}]}$ $\text{K}_1=\frac{[\text{B}]}{[\text{A}]},\text{K}_2=\frac{[\text{C}]}{[\text{B}]},$$\text{K}_3=\frac{[\text{D}]}{[\text{C}]}\text{ and }\text{K}_4=\frac{[\text{E}]}{[\text{D}]}$
$\because\text{K}=\text{K}_1\times\text{K}_2\times\text{K}_3\times\text{K}_4=\frac{[\text{E}]}{[\text{A}]}$ $=1\times2\times3\times4=24$

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Question 553 Marks

Write two equations to show the amphiprotic (acid as well as base) property of water.

Answer

$\begin{matrix}\text{HCl}&+&\text{H}_2\text{O}&\rightleftharpoons&\text{H}_3\text{O}^+&+&\text{Cl}^-\\\text{Acid}&&\text{Base}_2&&\text{Acid}_2&&\text{Base}_1\\\end{matrix}$
$\begin{matrix}\text{NH}_3&+&\text{H}_2\text{O}&\rightleftharpoons&\text{NH}^+_4&+&\text{OH}^-\\\text{Base}_1&&\text{Acid}_2&&\text{Acid}_1&&\text{Base}_2\end{matrix}$
In first equation $\mathrm{H}_2 \mathrm{O}$ acts as base and in second it acts as acid, so $\mathrm{H}_2 \mathrm{O}$ is amplification, i.e., can gain $\mathrm{H}^{+}$and lose $\mathrm{H}^{+}$.

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Question 563 Marks

Which of the following reactions involve homogeneous equilibrium and which involve heterogeneous equilibrium?

$\text{Ag}_2\text{O}(\text{s})+2\text{HNO}_3(\text{aq})\rightleftharpoons2\text{AgNO}_3(\text{aq})+\text{H}_2\text{O}(\text{l})$
$\text{C(s)}+\text{CO(}_2(\text{g})\rightleftharpoons2\text{CO(g)}$
$\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{H}_2\text{O}(\text{l})\\\rightleftharpoons\text{CH}_3\text{COOH}(\text{aq})+\text{C}_2\text{H}_5\text{OH}(\text{aq})$
$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$

Answer

Heterogeneous equilibrium.
Heterogeneous equilibrium.
Homogeneous equilibrium.
Homogeneous equilibrium.

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Question 573 Marks

Match the following graphical variation with their description.

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Chemistry 3 Marks Question