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Chemistry-

General

Easy

Question

An endothermic reaction with high activation energy for the forward reaction is given by the diagram:

The correct answer is:

Related Questions to study

Graph between concentration of the product' x' and time' t' for $A rightwards arrow B$ is given ahead:
The graph between $negative fraction numerator d left square bracket A right square bracket over denominator d t end fraction$ and time wil1 be of the type:

Graph between concentration of the product' x' and time' t' for $A rightwards arrow B$ is given ahead:
The graph between $negative fraction numerator d left square bracket A right square bracket over denominator d t end fraction$ and time wil1 be of the type:

Chemistry-General

Following is the graph between $left parenthesis a minus x right parenthesis to the power of negative 1 end exponent$ and time t for second order reaction. $theta equals t a n to the power of negative 1 end exponent invisible function application left parenthesis 1 divided by 2 right parenthesis semicolon O A equals 2 L m o l to the power of negative 1 end exponent$ hence rate at the start of reaction will be:

Following is the graph between $left parenthesis a minus x right parenthesis to the power of negative 1 end exponent$ and time t for second order reaction. $theta equals t a n to the power of negative 1 end exponent invisible function application left parenthesis 1 divided by 2 right parenthesis semicolon O A equals 2 L m o l to the power of negative 1 end exponent$ hence rate at the start of reaction will be:

Chemistry-General

Following is the graph between log tll2 and log a (a = initial concentration) for a given
reaction at $27 to the power of ring operator end exponent$ e. Hence, order is:

Following is the graph between log tll2 and log a (a = initial concentration) for a given
reaction at $27 to the power of ring operator end exponent$ e. Hence, order is:

Chemistry-General

parallel

For the reaction $N H subscript 4 end subscript superscript plus end superscript plus O C N ⟶ N H subscript 2 end subscript C O N H subscript 2 end subscript$ the probable mechanism is,
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell NH subscript 4 superscript plus plus OCN to the power of minus not stretchy rightwards harpoon over leftwards harpoon NH subscript 4 OCN end cell row cell text and end text space of 1em NH subscript 4 OCN not stretchy ⟶ NH subscript 2 CONH subscript 2 space of 1em text (slow) end text end cell end table$ The rate law will be:

For the reaction $N H subscript 4 end subscript superscript plus end superscript plus O C N ⟶ N H subscript 2 end subscript C O N H subscript 2 end subscript$ the probable mechanism is,
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell NH subscript 4 superscript plus plus OCN to the power of minus not stretchy rightwards harpoon over leftwards harpoon NH subscript 4 OCN end cell row cell text and end text space of 1em NH subscript 4 OCN not stretchy ⟶ NH subscript 2 CONH subscript 2 space of 1em text (slow) end text end cell end table$ The rate law will be:

Chemistry-General

A drop of solution (volume 0.05 mL) contains $3 cross times 10 to the power of negative 6 end exponent$ mole of W. If the rate constant of disappearance of H+ is 107 mol $text litre end text to the power of negative 1 end exponent sec to the power of negative 1 end exponent$ , how long would it take for W in the drop tell disappear?

A drop of solution (volume 0.05 mL) contains $3 cross times 10 to the power of negative 6 end exponent$ mole of W. If the rate constant of disappearance of H+ is 107 mol $text litre end text to the power of negative 1 end exponent sec to the power of negative 1 end exponent$ , how long would it take for W in the drop tell disappear?

Chemistry-General

P₄ + 5O₂ ¾ → X $Error converting from MathML to accessible text.$ Y, Y is

P₄ + 5O₂ ¾ → X $Error converting from MathML to accessible text.$ Y, Y is

Chemistry-General

parallel

Cu²⁺ + 2e^– ¾® Cu; log[Cu²⁺] vs E_red graph is of the type shown in figure where OA = 0.34V, then electrode potential of the half cell of Cu/Cu²⁺ (0.1 M) will be

Cu²⁺ + 2e^– ¾® Cu; log[Cu²⁺] vs E_red graph is of the type shown in figure where OA = 0.34V, then electrode potential of the half cell of Cu/Cu²⁺ (0.1 M) will be

Chemistry-General

Which statement is correct for alkaline earth metals

Which statement is correct for alkaline earth metals

Chemistry-General

The incorrect statement is

The incorrect statement is

Chemistry-General

parallel

Ketone can be prepared by

Ketone can be prepared by

Chemistry-General

Chemistry-General

Chemistry-General

parallel

Identify X in the sequence C₃H₉N $Error converting from MathML to accessible text.$ C₃H₈O $Error converting from MathML to accessible text.$ C₃H₆O₂

Identify X in the sequence C₃H₉N $Error converting from MathML to accessible text.$ C₃H₈O $Error converting from MathML to accessible text.$ C₃H₆O₂

Chemistry-General

Which will give tertiary butyl alcohol, when treated with Grignard’s reagent

Which will give tertiary butyl alcohol, when treated with Grignard’s reagent

Chemistry-General

A compound ‘A’ forms ‘B’ with sodium metal and again ‘A’ forms ‘C’ with PCl₅, but ‘B’ and ‘C’ form diethyl ether. Therefore ‘A’, ‘B’ and ‘C’ are respectively

A compound ‘A’ forms ‘B’ with sodium metal and again ‘A’ forms ‘C’ with PCl₅, but ‘B’ and ‘C’ form diethyl ether. Therefore ‘A’, ‘B’ and ‘C’ are respectively

Chemistry-General

parallel

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