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Byju's Answer

Standard X

Chemistry

Gay Lussac's Law

For reaction:...

Question

For reaction:
$3 F e (s) + 4 H_{2} O (g) ⇌ F e_{3} O_{4} + 4 H_{2} (g)$
at $1200 K$ it was observed that partial pressure of water vapours is $15$ torr and total pressure at equilibrium is $31$ torr the volume of $k_{p}$ for the raction at $1200 K$ is ?

9.7

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0.001

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1.3

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0.08

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Solution

The correct option is C $1.3$
$\begin{matrix} Given:- 3 F e + 4 H_{2} O g) ⇌ {F e}_{3} O_{4} + 4 H_{2} (g) at 1200 K \end{matrix}$
$\begin{matrix} P_{H_{2} O} & = 15 Torr P_{T o t a l} & = 31 Torr P_{Total} & = P_{H_{2} O} + P_{H_{2}} = 15 + P H_{2} O \Rightarrow P H_{2} O & = 31 - 15 = 16 Torr \end{matrix}$
$\begin{matrix} k p & = \frac{{(P H_{2})}_{2}^{4}}{{(P_{H 2 O})}_{H 2 O}^{4}} K_{p} & = \frac{(16)^{4}}{(15)^{4}} K p & = 1.295 K p & \approx 1.3 S_{0}, (c) & 1.3 is correct answer \end{matrix}$

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Similar questions

Q. For the equilibrium reaction

3 F e (s) + 4 H_{2} O (g) ⇋ F e_{3} O_{4} (s) + 4 H_{2} (g)

, the relationship between

K_{p}

and

K_{c}

is:

Q. Equilibrium constants are given (in atm) for the following reactions at

0^{o} C

S r C l_{2} . 6 H_{2} O_{(s)} ⇌ S r C l_{2} . 2 H_{2} O_{(s)} + 4 H_{2} O_{(g)},

K_{p} = 6.89 \times 10^{- 12}

The vapour pressure of water at

0^{o} C

is 4.58 torr.
Calculate the pressure of water vapour in equilibrium at

0^{o} C

NO and $B r_{2}$ at initial partial pressure of 98.4 and 41.3 torr, respectively were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of KP for the following reaction at 300K.

$2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g)$

N O

and

B r_{2}

at initial partial pressures of 98.4 and 41.3

t o r r

, respectively, were allowed to react at

300 K

. At equilibrium the total pressure was

110.5 t o r r

. Calculate the value of the equilibrium constant (

K p

) at

300 K

, for the reaction

2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g) .

Q. Equilibrium constants are given (in atm) for the following reactions at

0^{o} C

S r C l_{2} . 6 H_{2} O_{(s)} ⇌ S r C l_{2} . 2 H_{2} O_{(s)} + 4 H_{2} O_{(g)},

K_{p} = 6.89 \times 10^{- 12}

The vapour pressure of water at

0^{o} C

is 4.58 torr.
Calculate the pressure of water vapour in equilibrium at

0^{o} C

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For reaction:3Fe(s)+4H2O(g)⇌Fe3O4+4H2(g)at 1200K it was observed that partial pressure of water vapours is 15 torr and total pressure at equilibrium is 31 torr the volume of kp for the raction at 1200K is ?

The correct option is C 1.3 Given:- 3Fe+4H2Og)⇌Fe3O4+4H2( g) at 1200 KPH2O=15 Torr PTotal=31 Torr PTotal =PH2O+PH2=15+PH2O⇒PH2O=31−15=16 Torr kp=(PH2)4(PH2O)4Kp=(16)4(15)4Kp=1.295Kp≈1.3S0,(c)1.3 is correct answer

For reaction:
$3 F e (s) + 4 H_{2} O (g) ⇌ F e_{3} O_{4} + 4 H_{2} (g)$
at $1200 K$ it was observed that partial pressure of water vapours is $15$ torr and total pressure at equilibrium is $31$ torr the volume of $k_{p}$ for the raction at $1200 K$ is ?