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Allisa [31]
3 years ago
7

At 311 K, this reaction has a K c value of 0.0111 . X ( g ) + 2 Y ( g ) − ⇀ ↽ − 2 Z ( g ) Calculate K p at 311 K. Note that the

pressure is in units of atmosphere (atm).
Chemistry
1 answer:
Aleks [24]3 years ago
4 0

Answer:

K_{p}=4.35\times 10^{-4}

Explanation:

We know, K_{p}=K_{c}(RT)^{\Delta n}

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and \Delta n is difference in sum of stoichiometric coefficient of products and reactants

Here \Delta n=(2)-(2+1)=-1 and T = 311 K

So, K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}

Hence value of equilibrium constant in terms of partial pressure (K_{p}) is 4.35\times 10^{-4}

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