The average rate of the following reaction at a certain temperature is 9.04Ã—10-1 M/min. Calculate the...

3.9 Ratings (577 Votes)

H₂SO₄(aq) + Sr(NO₃)₂(aq) â†’ SrSO₄(s) + 2HNO₃(aq)

1Â Â Â Â :Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â : 1Â Â Â Â Â Â Â Â Â :Â Â 2

9.04*10^-1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â : 18.08 *10^ -1

Therefore chane in conc of HNO3 = 1.81 M/min

Moles changed of H2SO4 in one minÂ Â = 9.04 * 10^ -1

Â Â Â Â Â \"Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â in one secÂ Â = ( 9.04 * 10^ -1) / 60

Â Â Â Â Â \"Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â inn 34 secÂ Â =Â Â 9.04*10^ -1 * 34 / 60Â Â Â =.512 M

H₂SO₄(aq) + Sr(NO₃)₂(aq) â†’ SrSO₄(s) + 2HNO₃(aq)

1Â Â Â Â :Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â : 1Â Â Â Â Â Â Â Â Â :Â Â 2

.512 M Â Â Â Â Â Â Â Â Â .512Â Â M Â Â Â Â Â Â Â Â Â Â Â .512 M Â Â Â Â Â 1.024 M

So conc of H2SO4 after 34 seconds = .684-.512Â Â =Â Â 0.172 M

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