AT 27 DEGREE CELCIUS IT WAS OBSERVED -DURING A REACTION OF HYDROGENATION THAT THE PRESSURE OH H2 nbsp-GAS DECREASES FROM 2 ATM TO 1-1 ATM IN 75 MINUTES-CALCULATE THE RATE OF REACTION-MOLARITY-SEC-GIVEN R-0-082-LITRE ATOM K-1MOL-1

1. The reaction is a first order reaction as the rate units are Ms 1 (or mole L 1s 1). 2. nbsp;Thus,the rate equation : nbsp; nbsp; nbsp; k1 = nb ...

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Standard X

Chemistry

Rate of a Reaction

AT 27 DEGREE ...

Question

AT 27 DEGREE CELCIUS IT WAS OBSERVED ,DURING A REACTION OF HYDROGENATION THAT THE PRESSURE OH H₂GAS DECREASES FROM 2 ATM TO 1.1 ATM IN 75 MINUTES.CALCULATE THE RATE OF REACTION(MOLARITY/SEC).GIVEN R=0.082/LITRE ATOM K^-1MOL^-1

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Solution

1. The reaction is a first order reaction as the rate units are Ms^-1 (or mole L^-1s^-1).
2. Thus,the rate equation :

k₁ = $\frac{2.303}{t} \log (\frac{a}{a - x}) = \frac{2.303}{75 \times 60} \log \frac{2}{1.1} = 1.3 \times 10^{- 4} M s^{- 1}$

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AT 27 DEGREE CELCIUS IT WAS OBSERVED ,DURING A REACTION OF HYDROGENATION THAT THE PRESSURE OH H2 GAS DECREASES FROM 2 ATM TO 1.1 ATM IN 75 MINUTES.CALCULATE THE RATE OF REACTION(MOLARITY/SEC).GIVEN R=0.082/LITRE ATOM K-1MOL-1

1. The reaction is a first order reaction as the rate units are Ms-1 (or mole L-1s-1). 2. Thus,the rate equation : k1 = 2.303tlog(aa-x)=2.30375×60log21.1= 1.3 ×10-4 Ms-1

AT 27 DEGREE CELCIUS IT WAS OBSERVED ,DURING A REACTION OF HYDROGENATION THAT THE PRESSURE OH H₂GAS DECREASES FROM 2 ATM TO 1.1 ATM IN 75 MINUTES.CALCULATE THE RATE OF REACTION(MOLARITY/SEC).GIVEN R=0.082/LITRE ATOM K^-1MOL^-1

1. The reaction is a first order reaction as the rate units are Ms^-1 (or mole L^-1s^-1).
2. Thus,the rate equation :

k₁ = $\frac{2.303}{t} \log (\frac{a}{a - x}) = \frac{2.303}{75 \times 60} \log \frac{2}{1.1} = 1.3 \times 10^{- 4} M s^{- 1}$