NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI07:EQUILIBRIUM
314465
28g of \(\mathrm{N}_{2}\) and \(6 \mathrm{~g}\) of \(\mathrm{H}_{2}\) were mixed. At equilibrium \(17 \mathrm{~g}\) \(\mathrm{NH}_{3}\) was produced. The weights of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) at equilibrium are respectively
1 \(11 \mathrm{~g}, 0 \mathrm{~g}\)
2 \(1 \mathrm{~g}, 3 \mathrm{~g}\)
3 \(14 \mathrm{~g}, 3 \mathrm{~g}\)
4 11g,3g
Explanation:
\[\begin{array}{*{20}{c}} {}&{{{\text{N}}_2}}& + &{3{{\text{H}}_2}}& \rightleftharpoons &{2{\text{N}}{{\text{H}}_3}} \\ {{\text{Initially}}{\mkern 1mu} }&{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 28\;{\text{g}}}&{}&{6\;{\text{g}}}&{}&{0{\mkern 1mu} } \\ {}&{{\mkern 1mu} {\text{1 mole}}}&{}&{{\mkern 1mu} {\text{3 mole}}{\mkern 1mu} }&{}&{{\mkern 1mu} {\text{2 mole}}} \\ {{\text{At}}{\mkern 1mu} {\text{equil}}}&{(1 - 0.5)}&{}&{{\mkern 1mu} (3 - 1.5)}&{}&{17{\mkern 1mu} {\text{g}}{\mkern 1mu} \,{\text{N}}{{\text{H}}_{\text{3}}}} \end{array}\] Concentration of 1 mole of \(\mathrm{N}_{2}\) combines with 3 moles of \(\mathrm{H}_{2}\) to produce 2 moles of \(\mathrm{NH}_{3}\) but, since at equilibrium 1 mole of \(\mathrm{NH}_{3}\) is produced. So, 0.5 moles, i.e. \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) are left at equilibrium, Similarly, 3 moles of \(\mathrm{H}_{2} \equiv 2\) moles of \(\mathrm{NH}_{3}\). For 1 mole of \(\mathrm{NH}_{3} \equiv 1.5\) mole of \(\mathrm{H}_{2}\) Hence, 1.5 mole i.e. \(3 \mathrm{~g}\) of \(\mathrm{H}_{2}\) are left at equilibrium. So, \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) and 3g of \(\mathrm{H}_{2}\) are left at equilibrium.
CHXI07:EQUILIBRIUM
314466
4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is
1 6.5
2 7.0
3 8.0
4 2.0
Explanation:
Total number of moles \(\mathrm {=2+3+2=7}\) moles.
KCET - 2006
CHXI07:EQUILIBRIUM
314489
Assertion : The value of \(\mathrm{K}\) gives us a relative idea about the extent to which a reaction proceeds. Reason : The value of \(\mathrm{K}\) is independent of the stoichiometry of reactants and products at the point of equilibrium.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The equilibrium constant (K) remains constant at a given temperature but adjusts to changes in the reaction equation stoichiometry, with the value raised to the power of any factor by which the equation is multiplied. So option (3) is correct.
CHXI07:EQUILIBRIUM
314468
Three moles of \(\mathrm{PCl}_{5}\), three moles of \(\mathrm{PCl}_{3}\) and two moles of \(\mathrm{Cl}_{2}\) are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of \(\mathrm{PCl}_{5}\), the number of moles of \(\mathrm{PCl}_{3}\) present in it is
1 6
2 4.5
3 5
4 3
Explanation:
\[\begin{array}{*{20}{c}} {{\text{PC}}{{\text{l}}_{{\text{5(g)}}}} \rightleftharpoons }&{{\text{PC}}{{\text{l}}_{{\text{3(g)}}}}}&{\text{ + }}&{{\text{C}}{{\text{l}}_{{\text{2(g)}}}}} \\ {{\text{Initially:}}}&{\text{3}}&{\text{3}}&{\text{2}} \\ {{\text{At}}\,{\text{eqm:}}}&{{\text{(3 - x)}}}&{{\text{(3 + x)}}}&{{\text{(2 + x)}}} \end{array}\] Given that, \(\mathrm {(3-\mathrm{x})=1.5 \Rightarrow \mathrm{x}=1.5}\) Therefore no. of moles of \(\mathrm {\mathrm{PCl}_{3(\mathrm{~g})}}\) present is \(\rm { =(3+x) \\ =3+1.5=4.5}\)
314465
28g of \(\mathrm{N}_{2}\) and \(6 \mathrm{~g}\) of \(\mathrm{H}_{2}\) were mixed. At equilibrium \(17 \mathrm{~g}\) \(\mathrm{NH}_{3}\) was produced. The weights of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) at equilibrium are respectively
1 \(11 \mathrm{~g}, 0 \mathrm{~g}\)
2 \(1 \mathrm{~g}, 3 \mathrm{~g}\)
3 \(14 \mathrm{~g}, 3 \mathrm{~g}\)
4 11g,3g
Explanation:
\[\begin{array}{*{20}{c}} {}&{{{\text{N}}_2}}& + &{3{{\text{H}}_2}}& \rightleftharpoons &{2{\text{N}}{{\text{H}}_3}} \\ {{\text{Initially}}{\mkern 1mu} }&{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 28\;{\text{g}}}&{}&{6\;{\text{g}}}&{}&{0{\mkern 1mu} } \\ {}&{{\mkern 1mu} {\text{1 mole}}}&{}&{{\mkern 1mu} {\text{3 mole}}{\mkern 1mu} }&{}&{{\mkern 1mu} {\text{2 mole}}} \\ {{\text{At}}{\mkern 1mu} {\text{equil}}}&{(1 - 0.5)}&{}&{{\mkern 1mu} (3 - 1.5)}&{}&{17{\mkern 1mu} {\text{g}}{\mkern 1mu} \,{\text{N}}{{\text{H}}_{\text{3}}}} \end{array}\] Concentration of 1 mole of \(\mathrm{N}_{2}\) combines with 3 moles of \(\mathrm{H}_{2}\) to produce 2 moles of \(\mathrm{NH}_{3}\) but, since at equilibrium 1 mole of \(\mathrm{NH}_{3}\) is produced. So, 0.5 moles, i.e. \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) are left at equilibrium, Similarly, 3 moles of \(\mathrm{H}_{2} \equiv 2\) moles of \(\mathrm{NH}_{3}\). For 1 mole of \(\mathrm{NH}_{3} \equiv 1.5\) mole of \(\mathrm{H}_{2}\) Hence, 1.5 mole i.e. \(3 \mathrm{~g}\) of \(\mathrm{H}_{2}\) are left at equilibrium. So, \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) and 3g of \(\mathrm{H}_{2}\) are left at equilibrium.
CHXI07:EQUILIBRIUM
314466
4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is
1 6.5
2 7.0
3 8.0
4 2.0
Explanation:
Total number of moles \(\mathrm {=2+3+2=7}\) moles.
KCET - 2006
CHXI07:EQUILIBRIUM
314489
Assertion : The value of \(\mathrm{K}\) gives us a relative idea about the extent to which a reaction proceeds. Reason : The value of \(\mathrm{K}\) is independent of the stoichiometry of reactants and products at the point of equilibrium.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The equilibrium constant (K) remains constant at a given temperature but adjusts to changes in the reaction equation stoichiometry, with the value raised to the power of any factor by which the equation is multiplied. So option (3) is correct.
CHXI07:EQUILIBRIUM
314468
Three moles of \(\mathrm{PCl}_{5}\), three moles of \(\mathrm{PCl}_{3}\) and two moles of \(\mathrm{Cl}_{2}\) are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of \(\mathrm{PCl}_{5}\), the number of moles of \(\mathrm{PCl}_{3}\) present in it is
1 6
2 4.5
3 5
4 3
Explanation:
\[\begin{array}{*{20}{c}} {{\text{PC}}{{\text{l}}_{{\text{5(g)}}}} \rightleftharpoons }&{{\text{PC}}{{\text{l}}_{{\text{3(g)}}}}}&{\text{ + }}&{{\text{C}}{{\text{l}}_{{\text{2(g)}}}}} \\ {{\text{Initially:}}}&{\text{3}}&{\text{3}}&{\text{2}} \\ {{\text{At}}\,{\text{eqm:}}}&{{\text{(3 - x)}}}&{{\text{(3 + x)}}}&{{\text{(2 + x)}}} \end{array}\] Given that, \(\mathrm {(3-\mathrm{x})=1.5 \Rightarrow \mathrm{x}=1.5}\) Therefore no. of moles of \(\mathrm {\mathrm{PCl}_{3(\mathrm{~g})}}\) present is \(\rm { =(3+x) \\ =3+1.5=4.5}\)
314465
28g of \(\mathrm{N}_{2}\) and \(6 \mathrm{~g}\) of \(\mathrm{H}_{2}\) were mixed. At equilibrium \(17 \mathrm{~g}\) \(\mathrm{NH}_{3}\) was produced. The weights of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) at equilibrium are respectively
1 \(11 \mathrm{~g}, 0 \mathrm{~g}\)
2 \(1 \mathrm{~g}, 3 \mathrm{~g}\)
3 \(14 \mathrm{~g}, 3 \mathrm{~g}\)
4 11g,3g
Explanation:
\[\begin{array}{*{20}{c}} {}&{{{\text{N}}_2}}& + &{3{{\text{H}}_2}}& \rightleftharpoons &{2{\text{N}}{{\text{H}}_3}} \\ {{\text{Initially}}{\mkern 1mu} }&{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 28\;{\text{g}}}&{}&{6\;{\text{g}}}&{}&{0{\mkern 1mu} } \\ {}&{{\mkern 1mu} {\text{1 mole}}}&{}&{{\mkern 1mu} {\text{3 mole}}{\mkern 1mu} }&{}&{{\mkern 1mu} {\text{2 mole}}} \\ {{\text{At}}{\mkern 1mu} {\text{equil}}}&{(1 - 0.5)}&{}&{{\mkern 1mu} (3 - 1.5)}&{}&{17{\mkern 1mu} {\text{g}}{\mkern 1mu} \,{\text{N}}{{\text{H}}_{\text{3}}}} \end{array}\] Concentration of 1 mole of \(\mathrm{N}_{2}\) combines with 3 moles of \(\mathrm{H}_{2}\) to produce 2 moles of \(\mathrm{NH}_{3}\) but, since at equilibrium 1 mole of \(\mathrm{NH}_{3}\) is produced. So, 0.5 moles, i.e. \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) are left at equilibrium, Similarly, 3 moles of \(\mathrm{H}_{2} \equiv 2\) moles of \(\mathrm{NH}_{3}\). For 1 mole of \(\mathrm{NH}_{3} \equiv 1.5\) mole of \(\mathrm{H}_{2}\) Hence, 1.5 mole i.e. \(3 \mathrm{~g}\) of \(\mathrm{H}_{2}\) are left at equilibrium. So, \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) and 3g of \(\mathrm{H}_{2}\) are left at equilibrium.
CHXI07:EQUILIBRIUM
314466
4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is
1 6.5
2 7.0
3 8.0
4 2.0
Explanation:
Total number of moles \(\mathrm {=2+3+2=7}\) moles.
KCET - 2006
CHXI07:EQUILIBRIUM
314489
Assertion : The value of \(\mathrm{K}\) gives us a relative idea about the extent to which a reaction proceeds. Reason : The value of \(\mathrm{K}\) is independent of the stoichiometry of reactants and products at the point of equilibrium.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The equilibrium constant (K) remains constant at a given temperature but adjusts to changes in the reaction equation stoichiometry, with the value raised to the power of any factor by which the equation is multiplied. So option (3) is correct.
CHXI07:EQUILIBRIUM
314468
Three moles of \(\mathrm{PCl}_{5}\), three moles of \(\mathrm{PCl}_{3}\) and two moles of \(\mathrm{Cl}_{2}\) are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of \(\mathrm{PCl}_{5}\), the number of moles of \(\mathrm{PCl}_{3}\) present in it is
1 6
2 4.5
3 5
4 3
Explanation:
\[\begin{array}{*{20}{c}} {{\text{PC}}{{\text{l}}_{{\text{5(g)}}}} \rightleftharpoons }&{{\text{PC}}{{\text{l}}_{{\text{3(g)}}}}}&{\text{ + }}&{{\text{C}}{{\text{l}}_{{\text{2(g)}}}}} \\ {{\text{Initially:}}}&{\text{3}}&{\text{3}}&{\text{2}} \\ {{\text{At}}\,{\text{eqm:}}}&{{\text{(3 - x)}}}&{{\text{(3 + x)}}}&{{\text{(2 + x)}}} \end{array}\] Given that, \(\mathrm {(3-\mathrm{x})=1.5 \Rightarrow \mathrm{x}=1.5}\) Therefore no. of moles of \(\mathrm {\mathrm{PCl}_{3(\mathrm{~g})}}\) present is \(\rm { =(3+x) \\ =3+1.5=4.5}\)
314465
28g of \(\mathrm{N}_{2}\) and \(6 \mathrm{~g}\) of \(\mathrm{H}_{2}\) were mixed. At equilibrium \(17 \mathrm{~g}\) \(\mathrm{NH}_{3}\) was produced. The weights of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) at equilibrium are respectively
1 \(11 \mathrm{~g}, 0 \mathrm{~g}\)
2 \(1 \mathrm{~g}, 3 \mathrm{~g}\)
3 \(14 \mathrm{~g}, 3 \mathrm{~g}\)
4 11g,3g
Explanation:
\[\begin{array}{*{20}{c}} {}&{{{\text{N}}_2}}& + &{3{{\text{H}}_2}}& \rightleftharpoons &{2{\text{N}}{{\text{H}}_3}} \\ {{\text{Initially}}{\mkern 1mu} }&{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 28\;{\text{g}}}&{}&{6\;{\text{g}}}&{}&{0{\mkern 1mu} } \\ {}&{{\mkern 1mu} {\text{1 mole}}}&{}&{{\mkern 1mu} {\text{3 mole}}{\mkern 1mu} }&{}&{{\mkern 1mu} {\text{2 mole}}} \\ {{\text{At}}{\mkern 1mu} {\text{equil}}}&{(1 - 0.5)}&{}&{{\mkern 1mu} (3 - 1.5)}&{}&{17{\mkern 1mu} {\text{g}}{\mkern 1mu} \,{\text{N}}{{\text{H}}_{\text{3}}}} \end{array}\] Concentration of 1 mole of \(\mathrm{N}_{2}\) combines with 3 moles of \(\mathrm{H}_{2}\) to produce 2 moles of \(\mathrm{NH}_{3}\) but, since at equilibrium 1 mole of \(\mathrm{NH}_{3}\) is produced. So, 0.5 moles, i.e. \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) are left at equilibrium, Similarly, 3 moles of \(\mathrm{H}_{2} \equiv 2\) moles of \(\mathrm{NH}_{3}\). For 1 mole of \(\mathrm{NH}_{3} \equiv 1.5\) mole of \(\mathrm{H}_{2}\) Hence, 1.5 mole i.e. \(3 \mathrm{~g}\) of \(\mathrm{H}_{2}\) are left at equilibrium. So, \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) and 3g of \(\mathrm{H}_{2}\) are left at equilibrium.
CHXI07:EQUILIBRIUM
314466
4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is
1 6.5
2 7.0
3 8.0
4 2.0
Explanation:
Total number of moles \(\mathrm {=2+3+2=7}\) moles.
KCET - 2006
CHXI07:EQUILIBRIUM
314489
Assertion : The value of \(\mathrm{K}\) gives us a relative idea about the extent to which a reaction proceeds. Reason : The value of \(\mathrm{K}\) is independent of the stoichiometry of reactants and products at the point of equilibrium.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The equilibrium constant (K) remains constant at a given temperature but adjusts to changes in the reaction equation stoichiometry, with the value raised to the power of any factor by which the equation is multiplied. So option (3) is correct.
CHXI07:EQUILIBRIUM
314468
Three moles of \(\mathrm{PCl}_{5}\), three moles of \(\mathrm{PCl}_{3}\) and two moles of \(\mathrm{Cl}_{2}\) are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of \(\mathrm{PCl}_{5}\), the number of moles of \(\mathrm{PCl}_{3}\) present in it is
1 6
2 4.5
3 5
4 3
Explanation:
\[\begin{array}{*{20}{c}} {{\text{PC}}{{\text{l}}_{{\text{5(g)}}}} \rightleftharpoons }&{{\text{PC}}{{\text{l}}_{{\text{3(g)}}}}}&{\text{ + }}&{{\text{C}}{{\text{l}}_{{\text{2(g)}}}}} \\ {{\text{Initially:}}}&{\text{3}}&{\text{3}}&{\text{2}} \\ {{\text{At}}\,{\text{eqm:}}}&{{\text{(3 - x)}}}&{{\text{(3 + x)}}}&{{\text{(2 + x)}}} \end{array}\] Given that, \(\mathrm {(3-\mathrm{x})=1.5 \Rightarrow \mathrm{x}=1.5}\) Therefore no. of moles of \(\mathrm {\mathrm{PCl}_{3(\mathrm{~g})}}\) present is \(\rm { =(3+x) \\ =3+1.5=4.5}\)
