1
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
Which of the following combination will produce $${H_2}$$ gas ?
2
JEE Advanced 2017 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is (are)
3
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
For the following cell,
$$Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)$$
when the concentration of $$Z{n^{2 + }}$$ is $$10$$ times the concentration of $$C{u^{2 + }},$$ the expression for $$\Delta G$$ (in $$J\,mo{l^{ - 1}}$$) is [$$F$$ is Faraday constant; $$R$$ is gas constant; $$T$$ is temperature; $${E^0}$$ (cell)$$=1.1$$ $$V$$]
$$Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)$$
when the concentration of $$Z{n^{2 + }}$$ is $$10$$ times the concentration of $$C{u^{2 + }},$$ the expression for $$\Delta G$$ (in $$J\,mo{l^{ - 1}}$$) is [$$F$$ is Faraday constant; $$R$$ is gas constant; $$T$$ is temperature; $${E^0}$$ (cell)$$=1.1$$ $$V$$]
4
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
The standard state Gibbs free energies of formation of $$C$$(graphite) and $$C$$(diamond) at $$T=298$$ $$K$$ are
$${\Delta _f}{G^0}$$ [$$C$$(graphite)] $$ = 0kJmo{l^{ - 1}}$$
$${\Delta _f}{G^0}$$ [$$C$$(diamond)] $$ = 2.9kJmo{l^{ - 1}}$$
The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C$$(graphite)] to diamond [$$C$$(diamond)] reduces its volume by $$2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$$ If $$C$$(graphite) is converted to $$C$$(diamond) isothermally at $$T=298$$ $$K,$$ the pressure at which $$C$$(graphite) is in equilibrium with $$C$$(diamond), is
[Useful information : $$1$$ $$J=1$$ $$kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};$$ $$1$$ bar $$ = {10^5}$$ $$Pa$$]
$${\Delta _f}{G^0}$$ [$$C$$(graphite)] $$ = 0kJmo{l^{ - 1}}$$
$${\Delta _f}{G^0}$$ [$$C$$(diamond)] $$ = 2.9kJmo{l^{ - 1}}$$
The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C$$(graphite)] to diamond [$$C$$(diamond)] reduces its volume by $$2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$$ If $$C$$(graphite) is converted to $$C$$(diamond) isothermally at $$T=298$$ $$K,$$ the pressure at which $$C$$(graphite) is in equilibrium with $$C$$(diamond), is
[Useful information : $$1$$ $$J=1$$ $$kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};$$ $$1$$ bar $$ = {10^5}$$ $$Pa$$]
Paper analysis
Total Questions
Chemistry
18
Mathematics
18
Physics
18
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