For the reaction PCl-3-g-Cl-2-g-rightleftharpoons-PCl-5-g- the value of K-c 250-oC is 26- The value of K-p this temperature will be -0-610-570-830-46

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Toppr Admin · Accepted Answer

Kp-Kc-xD7-RT-x394-n-x394-n- no of moles of gaseous produt - number of moles of gaseous rectanttherefore -x394-n-1-x2212-2-x2212-1R - 0-0821- T - 250 - 273-15 - 523-15 and Kc - 26soKp-Kc-xD7-RT-x2212-1-xA0-Kp-260-0821-xD7-523-15-0-61

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 $K_{p}$ at this temperature will be :
0.61
0.57
0.83
0.46

$K_{p} = K_{c} \times (R T)^{Δ n}$
$Δ n =$ no of moles of gaseous produt - number of moles of gaseous rectant
therefore $Δ n = 1 - 2 = - 1$
R = 0.0821, T = 250 + 273.15 = 523.15 and $K_{c}$ = 26
so $K_{p} = K_{c} \times (R T)^{- 1}$
$K_{p} = \frac{26}{0.0821 \times 523.15} = 0.61$

For the reaction PCl3(g)+Cl2(g)⇌PCl5(g) the value of Kc at 250oC is 26. The value of Kp at this temperature will be :0.610.570.830.46

Kp=Kc×(RT)ΔnΔn= no of moles of gaseous produt - number of moles of gaseous rectanttherefore Δn=1−2=−1R = 0.0821, T = 250 + 273.15 = 523.15 and Kc = 26soKp=Kc×(RT)−1 Kp=260.0821×523.15=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 $K_{p}$ at this temperature will be :
0.61
0.57
0.83
0.46

$K_{p} = K_{c} \times (R T)^{Δ n}$
$Δ n =$ no of moles of gaseous produt - number of moles of gaseous rectant
therefore $Δ n = 1 - 2 = - 1$
R = 0.0821, T = 250 + 273.15 = 523.15 and $K_{c}$ = 26
so $K_{p} = K_{c} \times (R T)^{- 1}$
$K_{p} = \frac{26}{0.0821 \times 523.15} = 0.61$