Thermal decomposition of HCOOH is a first ord | vedclass

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(B) For a first order reaction,the rate equation is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $K = 4.606 	imes 10^{-3} \ s^{-1}

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$500$

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(B) For a first order reaction,the rate equation is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $K = 4.606 	imes 10^{-3} \ s^{-1}$ If $90 \%$ of the initial quantity decomposes,then the remaining quantity $[A]_t = 100 \% - 90 \% = 10 \%$. Let $[A]_0 = 100$. Then $[A]_t = 10$. Substituting the values: $t = \frac{2.303}{4.606 	imes 10^{-3}} \log \frac{100}{10}$ $t = \frac{2.303}{4.606 	imes 10^{-3}} \log(10)$ Since $\log(10) = 1$: $t = \frac{2.303}{4.606 	imes 10^{-3}} 	imes 1 = 0.5 	imes 10^3 = 500 \ s$.

Thermal decomposition of $HCOOH$ is a first order reaction and the rate constant at $T(K)$ is $4.606 \times 10^{-3} \ s^{-1}$. The time required to decompose $90 \%$ of initial quantity of $HCOOH$ at $T(K)$ in seconds is

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