What is the order of reaction $r\, = \,k{[A]^{\frac{3}{2}}}\,{[B]^2}$ ?
What is the order of reaction $r\, = \,k{[A]^{\frac{3}{2}}}\,{[B]^2}$ ?
Similar Questions
The rate of certain reaction depends on concentration according to the equation $\frac{{ - dc}}{{dt}}\, = \,\frac{{{K_1}C}}{{1 + {K_2}C}},$ what is the order, when concentration $(c)$ is very-very high
The rate of certain reaction depends on concentration according to the equation $\frac{{ - dc}}{{dt}}\, = \,\frac{{{K_1}C}}{{1 + {K_2}C}},$ what is the order, when concentration $(c)$ is very-very high
For the reaction $3\,{A_{\,(g)\,}}\,\xrightarrow{K}\,{B_{(g)}}\, + \,\,{C_{(g)\,,}}K$ is ${10^{ - 14}}\,L/mol.\min .$ if $(A) = 0.5\,M$ then the value of $ - \frac{{d(A)}}{{dt}}$ (in $M / sec$ ) is.
For the reaction $3\,{A_{\,(g)\,}}\,\xrightarrow{K}\,{B_{(g)}}\, + \,\,{C_{(g)\,,}}K$ is ${10^{ - 14}}\,L/mol.\min .$ if $(A) = 0.5\,M$ then the value of $ - \frac{{d(A)}}{{dt}}$ (in $M / sec$ ) is.
What is the order of reaction $A + B \to C$
Observation
$[A]$
$[B]$
Rate of reaction
$1$
$0.1$
$0.1$
$2 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$
$2$
$0.4$
$0.1$
$0.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$
$3$
$0.1$
$0.2$
$1.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$
What is the order of reaction $A + B \to C$
| Observation | $[A]$ | $[B]$ | Rate of reaction |
| $1$ | $0.1$ | $0.1$ | $2 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$ |
| $2$ | $0.4$ | $0.1$ | $0.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$ |
| $3$ | $0.1$ | $0.2$ | $1.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{\sec ^{ - 1}}$ |
For a reaction, $AB_5 \to AB + 4B$ The rate can be expressed in following ways
$\frac{{ - d[A{B_5}]}}{{dt}} = K[A{B_5}]$ ; $\frac{{d[B]}}{{dt}} = {K_1}[A{B_5}]$
So the correct relation between $K$ and $K_1$ is
For a reaction, $AB_5 \to AB + 4B$ The rate can be expressed in following ways
$\frac{{ - d[A{B_5}]}}{{dt}} = K[A{B_5}]$ ; $\frac{{d[B]}}{{dt}} = {K_1}[A{B_5}]$
So the correct relation between $K$ and $K_1$ is
Consider a reaction $\mathrm{aG}+\mathrm{bH} \rightarrow$ Products. When concentration of both the reactants $\mathrm{G}$ and $\mathrm{H}$ is doubled, the rate increases by eight times. However, when concentration of $\mathrm{G}$ is doubled keeping the concentration of $\mathrm{H}$ fixed, the rate is doubled. The overall order of the reaction is
Consider a reaction $\mathrm{aG}+\mathrm{bH} \rightarrow$ Products. When concentration of both the reactants $\mathrm{G}$ and $\mathrm{H}$ is doubled, the rate increases by eight times. However, when concentration of $\mathrm{G}$ is doubled keeping the concentration of $\mathrm{H}$ fixed, the rate is doubled. The overall order of the reaction is
- [IIT 2007]