The total pressure which Aleft- right- is50- dissociated according to the equation Aleft- right- Leftrightarrow Bleft- right- -Cleft- right- is numerically equal to-times of Kp-

Question

Toppr Admin · Accepted Answer

A-x21CC-B-C-Kp-PT-xD7-0-5R1-5RPT-xD7-0-51-5KP-PT-xD7-0-51-5KP-PT33-KP-PT-

The total pressure at which A $()$ is $50$ dissociated according to the equation A $() \Leftrightarrow B () + C ()$ is numerically equal to...........times of Kp.

$A ⇌ B + C () K_{p} = \frac{P_{T} \times \frac{0.5 R}{1.5 R}}{P_{T} \times \frac{0.5}{1.5}} K_{P} = P_{T} \times \frac{0.5}{1.5} K_{P} = \frac{P_{T}}{3} 3. K_{P} = P_{T} .$

The total pressure at which A()is50 dissociated according to the equation A()⇔B()+C()is numerically equal to...........times of Kp.

A⇌B+C()Kp=PT×0.5R1.5RPT×0.51.5KP=PT×0.51.5KP=PT33.KP=PT.

The total pressure at which A $()$ is $50$ dissociated according to the equation A $() \Leftrightarrow B () + C ()$ is numerically equal to...........times of Kp.

$A ⇌ B + C () K_{p} = \frac{P_{T} \times \frac{0.5 R}{1.5 R}}{P_{T} \times \frac{0.5}{1.5}} K_{P} = P_{T} \times \frac{0.5}{1.5} K_{P} = \frac{P_{T}}{3} 3. K_{P} = P_{T} .$