InterviewSolution
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InterviewSolution
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For the first order reaction `A(g) rarr 2B(g) + C(g)`, the initial presuure is `P_(A) = 90 m Hg`, the pressure after `10` minutes is found to be `180 mm Hg`. The rate constant of the reaction isA. `1.15 xx 10^(-3) sec^(-1)`B. `2.3 xx 10^(-3) sec^(-1)`C. `3.45 xx 10^(-3) sec^(-1)`D. `6 xx 10^(-3) sec^(-1)` |
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Answer» Correct Answer - A `{:("The reaction is",A_((g)),rarr,B_((g)),+, C_((g))),("at t=0 pressure",PA,,0,,0),("Pressure after",P_(4)-x,,2x,,x),("10 minutes",,,,,):}` Total pressure after `10` minutes, `P_(T) = P_(A) + P_(B) + P_( C)` or, `P_(T) = (P_(A) - x) + 2x + x` `= P_(A) + 2x = 180` `because P_(A) = 90mm` `90 + 2x = 180` or, `X = 45 mm` `:. P_(A) - x = 90 - 45 = 45 mm` Since, `kt = 2.303 "log"(P_(A))/(P_(A) - x)` or, `k xx 10 = 2.303 "log"(90)/(90 - 45)` `k xx 10 = 0.693` or, `k(0.693)/(10) = 0.0693 "min"^(-1)` `= (0.693)/(10 xx 60 sec) = 1.155 xx 10^(-3) sec^(-1)` |
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