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Question: If *K*<sub>p</sub> for reaction \(A_{(g)} + 2B_{(g)}\)⇌ \(3C_{(g)} + D_{(g)}\) is 0.05 *atm* at 1000...

If Kp for reaction A(g)+2B(g)A_{(g)} + 2B_{(g)}⇌ 3C(g)+D(g)3C_{(g)} + D_{(g)} is 0.05 atm at 1000K its Kc in term of R will be

A

5×10−4R\frac{5 \times 10^{- 4}}{R}

B

5R\frac{5}{R}

C

5×10−5R\frac{5 \times 10^{- 5}}{R}

D

None

Answer

5×10−5R\frac{5 \times 10^{- 5}}{R}

Explanation

Solution

Kp=Kc(RT)Δn⇒5×10−2=Kc(R×1000)1⇒Kc=5×10−5RK_{p} = K_{c}(RT)^{\Delta n} \Rightarrow 5 \times 10^{- 2} = K_{c}(R \times 1000)^{1} \Rightarrow K_{c} = \frac{5 \times 10^{- 5}}{R}