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6.1. Equilibrium - 1 (chemical Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.1. Equilibrium - 1 (chemical Equilibrium)1 Mark
MCQ
The standard free energy change $\left(\Delta G ^{\circ}\right)$ for $50 \%$ dissociation of $N _{2} O _{4}$ into $NO _{2}$ at $27^{\circ} C$ and $1\, atm$ pressure is $- x\, J \,mol ^{-1}$. The value of $x$ is $......$ (Nearest Integer)

[Given $: R =8.31 \,J \,K ^{-1} \,mol ^{-1}, \log 1.33=0.1239$ $\ln 10=2.3]$

  • A
    $520$
  • B
    $430$
  • C
    $931$
  • ✓
    $710$
✓

Answer

Correct option: D.
$710$
d
$\quad \quad\quad \quad N _{2} O _{4} \quad \rightleftharpoons \quad\quad2 NO _{2}$

$t =0 \quad\quad1 \,mol$

$t = t\quad \quad(1-0.5) \,mol \quad \quad 0.5 \times 2\,mol$

$\quad\quad\quad\quad=0.5 \,mol \quad\quad\quad 1\, mol$

$k _{ p }=\frac{\left(\frac{1}{1.5} \times 1\right)^{2}}{\left(\frac{0.5}{1.5} \times 1\right)}=\frac{1}{0.75}=\frac{100}{75}$

$=1.33$

$\Delta G ^{0}=- RT \ell nk _{ P }$

$=-8.31 \times 300 \times \ell n (1.33)=-710.45\, J / mol$

$=-710 \,J / mol$.

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