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v36.3...

v36.3

A

`1//2`

B

2

C

1

D

`1//4`

Text Solution

Verified by Experts

The correct Answer is:
B
`N_(2)" Moles "=(56)/(28)=2,`
`O_(2)" Moles="(128)/(32)=4, NO" Moles="(120)/(30)=4`
`N_(2)+O_(2) Leftrightarrow 2NO_(2)`
`P_(N_(2))=1 xx 2/10=0.2, P_(o_(2))=1 xx 4/10, P_(NO)=1 xx (4)/(10)=0.4`
`K_(P)=(0.4 xx 0.4)/(0.4 xx 0.2)=2`
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Thus emf of the cell = 0.91 V. Q.3.6. The cell in which the following reaction occurs : 2Fe^(3+) (aq) + 2I^(-)(aq) to 2Fe^(2+0 (aq) + I_(2)(s) has E_("cell")^(@) = 0.236 V at 298 K Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

3.6 + 36.6 + 3.66 + 0.36 + 3.0=?

Knowledge Check

  • If E_(ClO_(3)^(-)//ClO_(4)^(-))=-0.36 V & E_(ClO_(3)^(-)//ClO_(2)^(-))^(@)=0.33V at 300 K. The equilibrium concentration of perchlorate ion (ClO_(4)^(-)) which was initially 1.0 M in ClO_(3)^(-) when the reaction starts to attain the equilibrium, 2ClO_(3)^(-)hArr ClO_(2)^(-)+ClO_(4)^(-) Given : Anti log(0.509)=3.329

    A
    `0.0236M`
    B
    `0.0190M`
    C
    `0.123M`
    D
    `0.191M`

  • sqrt(3.6-0.36)=?

    A
    `1.8`
    B
    `3.24`
    C
    `10.4976`
    D
    none of these

  • If a=36 , then find the value of a^(36^@)-a^(0^36) .

    A
    36
    B
    0
    C
    1
    D
    35

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