The reaction quotient Q for :
`N_2(g)+3H_2(g) hArr 2NH_3(g)` is given by `Q=([NH_3]^2)/([N_2][H_2]^3)` The reaction will proceed in backward direction, when :
A. `QgtK_c`
B. `Q=0`
C. `Q=K_c`
D. `QltK_c`
1 Answers
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Which of the following is best suited to dry crystals of ammine salts (complexes in which `NH_3` is a ligand e.g., `[Ni(NH_3)_6]Br_2)`
A mixture of CaO and `NH_4Cl` is used to dry crystals of complexes containing `NH_3` as ligands. The water liberated by moist crystals is absorbed by CaO. This moisture helps...
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A concentration cell `Pt|H_2(g)|HCl(aq)||H_2SO_(4)(aq)|H_2(g)|Pt` is constructed using equal concentration of acids and equal number of moles of `H_2`
Correct Answer - BCD `underset(P_1)(H_2(g))|underset(C)(H^+)||underset(2C)(H^+)|underset(P_2)(H_2(g))` Anodic reaction `1/2 H_2 toH^+ +e` Cathodic reaction `H^+ +e to 1/2 H^2` `underset(("anode"))(1/2H_2)+underset(("cathode"))(H^+)tounderset(("cathode"))(1/2H_2)+underset(("anode"))(H^+)` `E_(cell)^@=0` `E_(cell)=E_(cell)^@-0.06/1"log"(sqrt((P_(H_2))_("cathode"))xx[H^+]_("anode"))/(sqrt((P_(H_2))_("anode"))xx[H^+]_("cathode"))` `E_(cell)=-0.06 "log"(sqrt1/9xxC/(2C))=-0.06 "log" 1/(2xx3)=0.06 ("log" 2+ log 3)=0.06(0.3+0.48)=0.0468` volt
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For the synthesis of ammonia at 300 K : `N_2(g)+3H_2(g)to2NH_3(g)` Calculate the value of `DeltaG^@` in Kcal and give your answer in magnitude by usin
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`NH_4COONH_2 (s) hArr 2NH_3(g)+CO_(2)(g)` If equilibrium pressure is 6 atm for the above reaction, `K_p` will be :
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1 mole each of `N_2(g)` and `O_2(g)` are introduced in a 1L evacuated vessel at 523 K and equilibrium `N_2(g)+O_(2)ghArr2NO(g)` is established.The con
Correct Answer - A,B,C,D `N_2(g)+O_2(g)hArr 2NO(g)` (A)For changing pressure volume has to be changed, though number of moles of NO(g) do not get changed but its concentration will get changed. (B)Temperature...
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Match Matrix `{:("Column I","Column II"),((A)Zn|Zn^(2+)( C)||Zn^(2+)(2C)|Zn,(p)"Spontaneous cell reaction"),((B)H_2(P)|HCl(1N)||H_2SO_4(1N)|H_2(P),(q)
Correct Answer - A`to`p,q,s ; B`to`r,s ; C`to`p,q ; D`to`p,q,s,t `E_("cell")=-0.0591/2"log"C/(2C)=8.89xx10^(-3)` Spontaneous, cell works, conc cell `E_("cell")=-0.0591 "log"1/0.5 =- 0.0591` Non spontaneous cell will not work conc cell `E_("cell")=-0.0591/2"log"0.01/0.1^2-0.74+0.8=0.06` Spontaneous cell...
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Given : `Zn(OH)_2(s) hArr Zn(OH)_2(aq) , K_1=10^(-6)` `Zn(OH)_2(aq) hArr [Zn(OH)]^(+) +OH^(-) , K_2=10^(-7)` `[Zn(OH)]^+(aq) hArr Zn^(+2)+OH^(-) , K_3
Correct Answer - 1 Dissolved `[Zn(OH)_2]=[Zn^(+2)]_(aq)+[Zn(OH)^(+)]_(aq)+(Zn(OH)_2)_(aq)+[Zn(OH)_3^(-)]+[Zn(OH)_4]^(2-)` Now, `[Zn(OH)_2]_(aq)=10^(-6)` M in saturated solution. so `[Zn(OH)]^(+)+(10^(-6)xx10^(-7))/([OH^-])=10^(-13)/([OH^(-)])` Similarly `[Zn^(+2)]=10^(-17)/([OH^(-)]^2),[Zn(OH)_3]^(-)=10^(-3) [OH^(-)]` `[Zn(OH)_4^(2-)]=K_5[Zn(OH)_3^(-)][OH^-]=(10^(-2)M^(-1))[OH^(-)]^(2)` Dissolved `Zn(OH)_2=10^(-17)/([OH^(-)]^(2))+10^(-13)([OH^(-)])+10^(-6)+10^(-3)[OH^(-)]+10^(-2)[OH^(-)]^2` `=10^(-17)/10^(-19)+10^(-13)/10^(-8)+10^(-6)+10^(-3)xx10^(-8)+10^(-18)=10^(-1)+10^(-5)+10^(-6)+10^(-11)=10^(-1)` -log `Zn(OH)_2(aq)`=1
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Assertion: `S_(N^(2))` reactions do not proceed with retention of configuration. Reason: `S_(N^(2))` reactions proceed in a single step.
Correct Answer - B Correct explanation. In `S_N^(2)` reactions, the nucleophile attacks the substrate from the side opposite to the side where the leaving group is present.
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Four numbers `n_1,n_2,n_3andn_4` are given as `n_1=sin15^@-cos15^@,n_2=cos93^@+sin93^@,n_3=tan27^@-cot27^@,n_4=cot127^@+tan127^@`,Then
Correct Answer - A::C::D `n_1=sin15^@-cos15^@lt-ve" "(cos15^@gtsin15^@)` `n_2=cos93^@+sin93^@` `=-sin3^@[email protected]" "(cos3^@gtsin3^@)` `n_3=tan27^@-cot27^@lt0" "(tan27^@[email protected])` `n_4=cot127^@+tan127^@lt0" "(tan127^@,cot127^@lt0)`.
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