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8.1 In Reaction 1, platinum (Pt) acts as a catalyst - NSC Physical Sciences - Question 8 - 2016 - Paper 2

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8.1 In Reaction 1, platinum (Pt) acts as a catalyst. What NAME is given to the energy that a catalyst changes in a chemical reaction? 8.2 Reaction 2 reaches equili... show full transcript

Worked Solution & Example Answer:8.1 In Reaction 1, platinum (Pt) acts as a catalyst - NSC Physical Sciences - Question 8 - 2016 - Paper 2

Step 1

What NAME is given to the energy that a catalyst changes in a chemical reaction?

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Answer

The energy that a catalyst changes in a chemical reaction is referred to as extbf{activation energy}.

Step 2

Is the reaction EXOTHERMIC or ENDOTHERMIC? Give a reason.

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Answer

The reaction is extbf{EXOTHERMIC} because the enthalpy change, ΔH, is negative (-149.1 kJ), indicating that energy is released during the reaction.

Step 3

Write down TWO changes that must be made to increase the YIELD of NO2.

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Answer

  1. extbf{Increase pressure} in the reaction system to favor the formation of fewer gas molecules.
  2. extbf{Add more NO or O2} to shift the equilibrium towards the production of NO2.

Step 4

What is the value of ΔH per mole of NO2 formed?

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Answer

The value of ΔH per mole of NO2 formed is extbf{-74.55 kJ}.

Step 5

Write down the NAME of the type of reaction between an acid and a base.

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Answer

The type of reaction between an acid and a base is called extbf{neutralisation} or extbf{protolysis}.

Step 6

Which particle (PROTON or ELECTRON) is transferred during the reaction mentioned in QUESTION 8.3.1?

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Answer

The particle transferred during the reaction is a extbf{PROTON}.

Step 7

Calculate the percentage purity of the ammonium nitrate sample.

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Answer

To calculate the percentage purity of the ammonium nitrate sample, first determine the number of moles of NH4NO3 that react:

n(NaOH)=2.4×103 moln(NaOH) = 2.4 \times 10^{-3} \text{ mol}

Since the mole ratio between NH4NO3 and NaOH is 1:1, the moles of NH4NO3 is also: n(NH4NO3)=2.4×103 moln(NH4NO3) = 2.4 \times 10^{-3} \text{ mol}

Now, calculate the mass of pure NH4NO3:

m(NH4NO3)=n(NH4NO3)×M(NH4NO3)=2.4×103 mol×80 g/mol=0.192 gm(NH4NO3) = n(NH4NO3) \times M(NH4NO3) = 2.4 \times 10^{-3} \text{ mol} \times 80 \text{ g/mol} = 0.192 \text{ g}

Now calculate the percentage purity: % Purity=0.192g0.204g×100%=94.12%\% \text{ Purity} = \frac{0.192 \, g}{0.204 \, g} \times 100 \% = 94.12 \%

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