For the reaction
$P {C l}_{3 (g)} + {C l}_{2 (g)} ⇌ P {C l}_{5 (g)}$
the value of $K_{c}$ at $250^{o} C$ is $26$ . The value of $K_{p}$ a this temperature will be

0.61

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.57

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.83

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.46

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.61$
$K_{p} = K_{c} {(R T)}^{Δ n_{g}}; Δ n_{g} = 1 - 2 = - 1$
$K_{p} = 26 {(0.082 \times 623)}^{- 1}$
$K_{p} = \frac{26}{0.082 \times 623} = 0.61$

Suggest Corrections

Similar questions

Q. For the following reaction at

250^{o} C

, the value of

K_{c}

26

, then the value of

K_{p}

at the same temperature will be:

P {C l}_{3} (g) + {C l}_{2} (g) ⇌ P {C l}_{5} (g)

For the reaction $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ at $2500^{\circ} C$ , the value of $K_{C}$ is $26$ , then the value of $K_{P}$ at the same temperature will be

Q. For the reaction

P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)

, the value of

K_{c}

250^{\circ} C

is 26. The value of

K_{p}

at the same temperature will be:

For the reaction $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ at $2500^{\circ} C$ , the value of $K_{C}$ is $26$ , then the value of $K_{P}$ at the same temperature will be

Q. For reaction

P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g), K_{c}

is 30 at 300 K. The value of

K_{p}

at 300 K is

Join BYJU'S Learning Program

For the reactionPCl3(g)+Cl2(g)⇌PCl5(g)the value of Kc at 250oC is 26. The value of Kp a this temperature will be

The correct option is B 0.61Kp=Kc(RT)Δng;Δng=1−2=−1Kp=26(0.082×623)−1Kp=260.082×623=0.61

For the reaction
$P {C l}_{3 (g)} + {C l}_{2 (g)} ⇌ P {C l}_{5 (g)}$
the value of $K_{c}$ at $250^{o} C$ is $26$ . The value of $K_{p}$ a this temperature will be

The correct option is B $0.61$
$K_{p} = K_{c} {(R T)}^{Δ n_{g}}; Δ n_{g} = 1 - 2 = - 1$
$K_{p} = 26 {(0.082 \times 623)}^{- 1}$
$K_{p} = \frac{26}{0.082 \times 623} = 0.61$