Question

For reaction: A(s)⇌B(g)+C(g). What will be the value of natural logarithm of ratio of total pressure at 400 K to that at 300 K[=ln(P300​/P400)​​] if △H=16.628 kJ. [Given: R=8.314 J/K−mole].

Answer 1

Answer:

1

Explanation:

1

10

1.0×10−10

1.0×1010

Answer :

A

Solution :

ΔG∘=ΔH∘−T.ΔS∘

=−29.8+298×(0.1)

=−29.8+29.8

∵ΔG∘=0

apply relation between

ΔG∘=−RTlnKeq

∴Keq=1

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Answer 2

Answer:

5/6

Explanation:

A(s) ⇌ B(g) + C(g)

Δ$n_{g}$ = 2 - 0 = 2

∴ $K_{p}$ = $K_{x}$ . $P^{n_{g} }$

where $K_{x}$ is the equilibrium constant of mole fraction

& P is total pressure

$K_{1}$ = $K_{x}$ * $P_{1} ^{2}$

$K_{2}$ = $K_{x}$ * $P_{2} ^_{2}$

∴ ln ( $K_{2}$ / $K_{1}$ ) = ΔH° / R ( $\frac{1}{T_{1} }$ - $\frac{1}{T_{2} }$ )

$ln$ ( $\frac{K_{x} . P_{1} ^{2}}{K_{x} . P_{2} ^{2}}$ ) = $\frac{16628}{8.314}$ ( $\frac{1}{300}$ - $\frac{1}{400}$ )

  $ln$ $($ $\frac{P_{1} }{P_{2}}$  $)^{2}$ = $2000$ ( $\frac{1}{1200}$ )

  $2$ $ln$ $($ $\frac{P_{1} }{P_{2}}$ $)$ =  $\frac{20}{12}$

     $ln$ $($ $\frac{P_{1} }{P_{2}}$ $)$ = $\frac{5}{6}$

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