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2 years ago

State what would be observed when the following pairs of reagents are mixed in a test tube.

Chemistry

1 answer:

Mandarinka [93]2 years ago

4 0

Answer:

(i). C6H2COOH and Na2CO3(aq)

observation: Bubbles of a colourless gas (carbon dioxide gas)

(ii) CH3CH2CH2OH and KMnO4 /H

observation: The orange solution turns green.

[This is because oxidation of propanol to propanoic acid occurs]

(iii) CH3CH2OH and CH3COOH + conc. H2SO4

observation: A sweet fruity smell is formed.

[This is because an ester, diethylether is formed]

(iv) CH3CH = CHCH3 and Br2 /H2O

observation: a brown solution is formed.

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$\large \boxed{\text{-41.2 kJ/mol}}$

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$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

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