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Standard XII

Chemistry

Rate Constant

If no catalys...

Question

If no catalyst $(H^{+})$ is present in acid hydrolysis of ester then constant k is :

\frac{2.303}{t} log \frac{V_{0}}{(V_{t} - V_{\infty})}

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\frac{2.303}{t} log \frac{V_{\infty}}{(V_{\infty} - V_{t})}

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\frac{2.303}{t} log \frac{V_{0}}{V_{t}}

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\frac{2.303}{t} log \frac{V_{\infty}}{(V_{t} - V_{\infty})}

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Solution

The correct option is A $\frac{2.303}{t} log \frac{V_{0}}{(V_{t} - V_{\infty})}$
In acid hydrolysis of ester,
$k = \frac{2.303}{t} log (\frac{V_{0}}{V_{t} - V_{\infty}})$ $V_{0} =$ Initial case, $V_{t} - V_{\infty} =$ final case.

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Similar questions

Q. Acid hydrolysis of ester in first order reaction and rate constant is given by

k = \frac{2.303}{t} l o g \frac{V_{\infty} - V_{0}}{V_{\infty} - V_{t}}

where

V_{0}

V_{t}

and

V_{\infty}

are the volume of standard

N a O H

required to neutralise acid present at a given time, if ester is

50

neutralised then:

Q. Acid hydrolysis of ester is first order reaction and rate constant is given by

k = \frac{2.303}{t} log \frac{V_{\infty} - V_{0}}{V_{\infty} - V_{t}}

where,

V_{0}

V_{t}

and

V_{\infty}

are the volume of standard

N a O H

required to neutralise acid present at a given time, if ester is

50

% neutralised then :

Q. In a device

I = 40 μ A [e^{V / V_{T}} - 1]

where

V_{T} = 0.025 V

. Find the dynamic resistance when

V = 0.5 V

$A (g) \to B (g) + C (g)$

- $\frac{d [A]}{d t} = k [A]$

At the beginning of this reaction, the pressure is 100 mm Hg. After 10 min, pressure increases to 120 mm Hg. The rate constant $(m i n^{- 1})$ is:

If [A] is the concentration of reactant A at any time t and [A_o] is the concentration at t=0, then for the first order reaction the rate equation can be written as:

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If no catalyst (H+) is present in acid hydrolysis of ester then constant k is :

The correct option is A 2.303tlogV0(Vt−V∞)In acid hydrolysis of ester,k=2.303tlog(V0Vt−V∞) V0= Initial case,Vt−V∞= final case.

If no catalyst $(H^{+})$ is present in acid hydrolysis of ester then constant k is :

The correct option is A $\frac{2.303}{t} log \frac{V_{0}}{(V_{t} - V_{\infty})}$
In acid hydrolysis of ester,
$k = \frac{2.303}{t} log (\frac{V_{0}}{V_{t} - V_{\infty}})$ $V_{0} =$ Initial case, $V_{t} - V_{\infty} =$ final case.