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MrRissso [65]
3 years ago
11

HELPPPyall please help my sister won't help me ​

Chemistry
1 answer:
ira [324]3 years ago
3 0

\text{H}_3\text{PO}_4  + 3\text{KOH}  \longrightarrow \text{K}_3 \text{PO}_4     +3\text{H}_2 \text{O}\\\\10\text{Na} +2 \text{NaNO}_3  \longrightarrow 6\text{Na}_2 \text{O} + \text{N}_2\\\\\  2\text{H}_3 \text{PO}_4 + 3\text{Mg(OH)}_2 \longrightarrow \text{Mg}_3(\text{PO}_4)_2+6\text{H}_2 \text{O}\\\\2\text{Al(OH)}_3 + 3\text{H}_2 \text{CO}_3 \longrightarrow \text{Al}_2(\text{CO}_3)_3 + 6\text{H}_2 \text{O} \\\\2\text{C}_6\text{H}_6+15\text{O}_2 \longrightarrow 12\text{CO}_2 +6\text{H}_2 \text{O}\\

\text{Pb(OH)}_2 +2\text{HCl} \longrightarrow \text{PbCl}_2 +2 \text{H}_2 \text{O}

2\text{Mn} + 6\text{HI} \longrightarrow 3 \text{H}_2+2\text{MnI}_3

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Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energ
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<u>Answer:</u> The Gibbs free energy of the reaction is 21.32 kJ/mol

<u>Explanation:</u>

The chemical equation follows:

\text{Malate }+NAD^+\rightleftharpoons \text{Oxaloacetate }+NADH

The equation used to Gibbs free energy of the reaction follows:

\Delta G=\Delta G^o+RT\ln K_{eq}

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K_{eq} = Ratio of concentration of products and reactants = \frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}

\text{[Oxaloacetate]}=0.130mM

[NADH]=2.0\times 10^2mM

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Putting values in above expression, we get:

\Delta G=29700J/mol+(8.314J/K.mol\times 310K\times \ln (\frac{0.130\times 2.0\times 10^2}{1.37\times 490}))\\\\\Delta G=21320.7J/mol=21.32kJ/mol

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol

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