Question

For the reaction:

SrCo3(s) \rightleftharpoons SrO(s)+CO2(s)

$\sf \: the \: value\: of equilibrium \: \\ \sf constant \: K_{p}=2.2×10^{-4} \: \\ \sf at\:1002K. Calculate \: K_{C} \: for \: the\: \\ \sf reaction.$​

Answer 1

Answer:

2.4 × 10 ⁻⁶  

Explanation:

Kp = Kc ( RT)Δⁿ

Δn = moles of gaseous products - moles of gaseous reactants

    = 1 - 0 = 1

Kc = Kp/RT

    = 2.2 × 10⁻⁴/ 0.0831 × 1002

    =  2.2 × 10⁻⁴/ 83.16

     = 2.4 × 10 ⁻⁶