Answer:
[tex] \bf{1.2 \times {10}^{-3}} [/tex] (Option A).
Explanation:
Given: [tex] \sf{{K}_{c} = 5.0 \times {10}^{-2}} [/tex]
Temperature (T) = 500K
Universal Gas Constant(R) = [tex] \sf{0.0821 {dm}^{3} atm{K}^{-1} {mol}^{-1} = 0.0821 L atm{K}^{-1} {mol}^{-1} } [/tex]
Here the reaction is:
[tex] \sf{2{SO}_{2} (g) + {O}_{2} (g) ⇌ 2 {SO}_{3} (g)} [/tex]
∆n = moles of product gas - moles of reactant gas = 2-3 = -1 mol gas.
Then, [tex] \sf{{K}_{p} = {K}_{c}{(RT)}^{∆ \: n}} [/tex]
[tex] \sf{{K}_{p} = 5.0 \times {10}^{-2} \times {(0.0821 L atm{K}^{-1} {mol}^{-1} × 500K)}^{-1}} [/tex]
[tex] \bf{ = 1.2 \times {10}^{-3}} [/tex]
[tex] \rule{230pt}{3pt} [/tex]
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Answers & Comments
Answer:
[tex] \bf{1.2 \times {10}^{-3}} [/tex] (Option A).
Explanation:
Given: [tex] \sf{{K}_{c} = 5.0 \times {10}^{-2}} [/tex]
Temperature (T) = 500K
Universal Gas Constant(R) = [tex] \sf{0.0821 {dm}^{3} atm{K}^{-1} {mol}^{-1} = 0.0821 L atm{K}^{-1} {mol}^{-1} } [/tex]
Here the reaction is:
[tex] \sf{2{SO}_{2} (g) + {O}_{2} (g) ⇌ 2 {SO}_{3} (g)} [/tex]
∆n = moles of product gas - moles of reactant gas = 2-3 = -1 mol gas.
Then, [tex] \sf{{K}_{p} = {K}_{c}{(RT)}^{∆ \: n}} [/tex]
[tex] \sf{{K}_{p} = 5.0 \times {10}^{-2} \times {(0.0821 L atm{K}^{-1} {mol}^{-1} × 500K)}^{-1}} [/tex]
[tex] \bf{ = 1.2 \times {10}^{-3}} [/tex]
[tex] \rule{230pt}{3pt} [/tex]