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Question: Inversion of sucrose \((C_{\text{12}}H_{\text{22}}O_{\text{11}})\) is first-order reaction and is st...

Inversion of sucrose (C12H22O11)(C_{\text{12}}H_{\text{22}}O_{\text{11}}) is first-order reaction and is studied by measuring angle of rotation at different instant of time

}{Sucrose Glu\cos eFructose}$$ If $(r_{\infty}\text{- }\text{r}_{0}) = \text{a and }(r_{\infty}\text{- }\text{r}_{t}) = (\text{a - x})\ $(where $r_{0}\text{ , }\text{r}_{t}$ and $r_{\infty}$ are the angle of rotation at the start, at the time t and at the end of the reaction respectively, then there is 50% inversion when :
A

r0 = 2rt - r∞r_{0}\text{ = 2}\text{r}_{t}\text{ - }\text{r}_{\infty}

B

r0 = rt - r∞r_{0}\text{ = }\text{r}_{t}\text{ - }\text{r}_{\infty}

C

r0 = rt - 2r∞r_{0}\text{ = }\text{r}_{t}\text{ - 2}\text{r}_{\infty}

D

r0 = rt + r∞r_{0}\text{ = }\text{r}_{t}\text{ + }\text{r}_{\infty}

Answer

r0 = 2rt - r∞r_{0}\text{ = 2}\text{r}_{t}\text{ - }\text{r}_{\infty}

Explanation

Solution

Given (r∞ - r0) = a ,(r∞ - rt) = (a - x)(r_{\infty}\text{ - }\text{r}_{0})\text{ = a ,} (r_{\infty}\text{ - }\text{r}_{t})\text{ = }(\text{a - x})

At 50% Inversion

a2=(a−x)\frac{a}{2} = (a - x)

(r∞−r0)2=(r∞−r0)\frac{\left( r_{\infty} - r_{0} \right)}{2} = \left( r_{\infty} - r_{0} \right)

}{r_{0}\text{ = 2}\text{r}_{t}\text{ - }\text{r}_{\infty}\ }$$