Explanation:
Experiment Initial [CS2] (mol/L) Initial Rate (mol/L·s)
1 0.100 2.7 × 10−7
2 0.080 2.2 × 10−7
3 0.055 1.5 × 10−7
4 0.044 1.2 × 10−7
a) Choose the rate law for the decomposition of CS2.
Comparing equations 1 and 3, reducing the initial concentration by almost half (from 0.100 to 0.055) leads too the rate of reaction to be reduced by almost half (from 2.7 × 10−7 to 1.5 × 10−7).
This signifies that the reaction is a first order reaction.
Rate = k [CS2]
(b) Calculate the average value of the rate constant.
Taking equation 1.
Rate = k [CS2]
k = Rate / [CS2]
k = 0.100 / (2.7 × 10−7) = 0.037 x 10^8 = 3.7 x 10^6s-1
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Answer:
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Explanation:
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1) Balanced chemical equation:
<span>C(s)+O2(g)→CO2(g), δh∘rxn=−393.5kj
2) Meaning: When burned 1 mol of C(s), this is solid pure charcoal, produces 1 mol of CO2 and liberates 393.5 kJ of heat
Ratio: 1 mol CO2 : 393.5 kJ
3) Proportion:
1 mol CO2 x
--------------- = -------------------
393.5 kJ 4.6 * 10^2 kJ
4) Solve for x:
x = 460 kJ * 1 mol CO2 / 393.5 kJ = 1.1690 mol CO2
5) convert moles to grams
mass in grams = number of moles * molar mass
mass in grams = 1.1690 mol * 44.01 g / mol = 51.4 g
Answer: 51.4 g
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