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JEE CHEMISTRY MCQ: Equilibrium

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An amount of 3 moles of N2\mathrm{N}_2 and some H2\mathrm{H}_2 is introduced into an evacuated vessel. The reaction starts at t=0t=0 and equilibrium is attained at t=t1t=t_1. The amount of ammonia at t=2t1t=2 t_1 is found to be 34 g.34 \mathrm{~g}. It is observed that w(N2)w(H2)=143\frac{w\left(\mathrm{N}_2\right)}{w\left(\mathrm{H}_2\right)}=\frac{14}{3} at t=t13t=\frac{t_1}{3} and t=t12.t=\frac{t_1}{2}. The only correct statement is
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Correct option: B
w(N2)+w(H2)+w(NH3)=102w\left(\mathrm{N}_2\right)+w\left(\mathrm{H}_2\right)+w\left(\mathrm{NH}_3\right)=102 g at t=2t1t=2 t_1
Option B (w(N2)+w(H2)+w(NH3)=102w(N_2) + w(H_2) + w(NH_3) = 102 g at t=2t1t=2t_1) is correct because the total mass remains constant throughout the reaction, and the initial mass is calculable from the given moles of N2N_2 and the stoichiometric amount of H2H_2.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is: K1\mathrm{K}_1 and K2\mathrm{K}_2 are equilibrium constant for reactions (i) and (ii)

N2( g)+O2( g)⇌2NO(g)…\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots (i)

NO(g)⇌12 N2( g)+12O2( g)…\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots (ii)

Then, Ka1,   Ka2   and   Ka3K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3} are the respective ionisation constants for the following reactions
H2S⇋H++HS−HS−⇋H++S2−H2S⇋2H++S2−\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}
The correct relationship between Ka1,  Ka2 ,Ka3K{a_1},\,\,K{a_2}\,,K{a_3} is The pH\mathrm{pH} of 0.1 M−N2H40.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4 solution is (For N2H4,\mathrm{N}_2 \mathrm{H}_4, Kb1=3.6×10−6,Kb2=6.4×10−12,log⁡2=0.3K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3, log⁡3=0.48\log 3=0.48) Reaction BaO2(s)⇌BaO(s)+O2(g);ΔH=−ve.{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}. In equilibrium condition pressure of O2\text{O}_2 depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.
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